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3 years ago

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1) Elevii unei clase au schimbat intre ei fotografii , astfel ca fiecare elev a primit cate o fotografie de la toti colegii sai.

Aratati ca numarul de fotografii este par. Creterii de divibilitate cu 2,5 si 10^n 2) Fie numerele 42,688,32,15,126,1114,71,3012,455,125,112,60,14,751,10,20,45,300,126,500,520,1700,430. Scrieti numerele care sunt divizibile cu . a)2 b)5 c)10 3) Care este cel mai mare numar natural de patru cifre diferite divizibile cu 2? Dar cu 5? Dar cu 2 si 5 ? Urgent

1 answer:

Schach [20]3 years ago

8 0

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What is the square root of 25

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3 years ago

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Ted earned $4,300 from his summer internship that he plans to invest so that he can buy a used car after three years. Ted choose

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Step-by-step explanation:

To find out if Ted can afford the car, find the future value of $4,300 in three years.

First convert the number of years and rates to quarterly values as this is the compounding period:

Term = 3 * 4 quarters = 12 quarters

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2 years ago

Evaluate 3 to the second power + (8 - 2 ) • 4 - 6/3

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2 years ago

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.

So

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

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

From what we have in a) and b)

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

So

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

(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:

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

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3 years ago

Given 43-x = 21, find the value of x

Vesnalui [34]

Answer:

X=22

Step-by-step explanation:

hope this helps, please mark brainliest

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2 years ago

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