What is the first step when applying properties of operations to divide 73.85 by 4.1?

The Brown's Ferry incident of 1975 focused national attention on the ever-present danger of fires breaking out in nuclear power plants. The Nuclear Regulatory Commission has estimated that with present technology there will be on average, one fire for every 10 years for a reactor. Suppose that a certain state has two reactors on line in 2020 and they behave independently of one another. Assuming the incident of fires for individual reactors can be described by a Poisson distribution, what is the probability that by 2030 at least two fires will have occurred at these reactors?

Answer:

The value is $P(x_1 + x_2 \ge 2 )= 0.5940$

Step-by-step explanation:

From the question we are told that

The rate at which fire breaks out every 10 years is $\lambda = 1$

Generally the probability distribution function for Poisson distribution is mathematically represented as

$P(x) = \frac{\lambda^x}{ k! } * e^{-\lambda}$

Here x represent the number of state which is 2 i.e $x_1 \ \ and \ \ x_2$

Generally the probability that by 2030 at least two fires will have occurred at these reactors is mathematically represented as

$P(x_1 + x_2 \ge 2 ) = 1 - P(x_1 + x_2 \le 1 )$

=> $P(x_1 + x_2 \ge 2 ) = 1 - [P(x_1 + x_2 = 0 ) + P( x_1 + x_2 = 1 )]$

=> $P(x_1 + x_2 \ge 2 ) = 1 - [ P(x_1 = 0 , x_2 = 0 ) + P( x_1 = 0 , x_2 = 1 ) + P(x_1 , x_2 = 0)]$

=> $P(x_1 + x_2 \ge 2 ) = 1 - P(x_1 = 0)P(x_2 = 0 ) + P( x_1 = 0 ) P( x_2 = 1 )+ P(x_1 = 1 )P(x_2 = 0)$

=> $P(x_1 + x_2 \ge 2 ) = 1 - \{ [ \frac{1^0}{ 0! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]] )+ ( [ \frac{1^1}{1! } * e^{-1}] * [[ \frac{1^1}{ 1! } * e^{-1}]] ) + ( [ \frac{1^1}{ 1! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]]) \}$

=> $P(x_1 + x_2 \ge 2 )= 1- [[0.3678 * 0.3679] + [0.3678 * 0.3679] + [0.3678 * 0.3679] ]$

$P(x_1 + x_2 \ge 2 )= 0.5940$

3 0

3 years ago

2+8<br><br> PLSSS HELP PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS

romanna [79]

Answer:

10

Step-by-step explanation

2 + 8= 10

6 0

3 years ago

Read 2 more answers

Will this happen in avenger 4?

Marizza181 [45]

In chess, the Endgame is where you sacrafice pawns, or in this case, minor characters in order to get a powerful piece back. You can leave some pawns in battle while regaining power pieces. The pawns sacraficed were Peter, Stephen, Bucky, Drax, T'challa. Mantis, etc. While leaving behind two pawns: Nebula, and Bruce Banner. Normally I like to think of the Hulk as a Rook, but since he's completely useless at the moment, he's a pawn. Nebula's fairly worthless, so she's a pawn. Thanos is playing with all power pieces and one pawn: Gamora. He sacraficed his pawn in order to complete his queen equivilence: the gauntlet. Now he's playing with all power pieces, while the Avenger's have sacraficed their pawns in order to get their queen: Captain Marvel, who in turn will wage war on Thanos only to find that a pawn has made it across the board and turned into the Hulk, and fights side by side the original Avengers to get the soul stone, revive Tony, who probably dies, get their friends home, welcome new friends, and kill Thanos.

Sorry about the rambling. I'm not even sure if I got to the point.

4 0

3 years ago