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3 years ago

6

An advertiser drops 10,000 leaflets on a city which has 2000 blocks. Assume that each leaflet has an equal chance of landing on

each block. What is the probability that a particular block will receive no leaflets

1 answer:

Ira Lisetskai [31]3 years ago

5 0

Answer:

0.00673

Step-by-step explanation:

As each of the 10000 leaflets has equal chance on landing into 1 of the 2000 blocks, this means each leaflet has a probability of 1/2000 to land on the 1st block, 1/2000 to land on the 2nd lock, 1/2000 to land on the 3rd block, and so on.

The probability that a particular block will receive no leaflet is chance of each leaflets to land on the other 1999 blocks, which has a chance of 1999/2000. For all 10000 leaflets to happen like this independently, then the total probability is

$\left(\frac{1999}{2000}\right)^{10000} = 0.00673$

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tekilochka [14]

Answer:

5x + y = -48

Step-by-step explanation:

(y + 3)/(x + 9) =m

=> y + 3 = -5x -45

=> 5x + y + 48 = 0

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There are 3 people pulling on an object, the first person pulls with a force of 226 newtons at a direction of 145 degrees the se

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Answer:

$R=883.74N$

Step-by-step explanation:

From the question we are told that:

Force 1 $F_1=226N$

Force 2 $F_2=331N$

Force 3 $F_3=377N$

Direction 1 $\theta_1=145 degrees \approx 35 \textdegree (2nd qudrant)$

Direction 2 $\theta_2=303 degrees \approx 57 \textdegree (3rd qudrant)$

Direction 3 $\theta_3=77 degrees$

Generally Resolving forces to X axis is mathematically given by

$\sum f_x=337cos77+331cos57+226sin35$

$\sum f_x=351.596N$

Generally Resolving forces to Y axis is mathematically given by

$\sum f_y=337sin77+331sin57+226cos35$

$\sum f_x=810.788N$

Generally the equation for Resultant force R is mathematically given by

$R=\sqrt{\sum f_x^2+\sum f_x^2}$

$R=\sqrt{351.596^2+810.788^2}$

$R=883.74N$

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3 years ago

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

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

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

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

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

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

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Answer:

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