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W + 19 = 49 and W = 30<br> O True<br> False

Illusion [34]

True because 30+19=49

5 0

2 years ago

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5x + 2 – 2 = -18 – 2 5x = -20 true or false

Oksi-84 [34.3K]

If 5x=-20 then it would be
$- 20 + 2 - 2 = - 18 - 2$
adding 2 and subtracting 2 they cancel out so its
$- 20 = - 18 - 2$
and if you subtract 2 from negative 18 it is negative 20 so it is true

5 0

3 years ago

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A skilled chess player believes that when they play a novice opponent, there is a 90% probability they will be able to beat them

zavuch27 [327]

Answer:

b) Binomial

c) Poisson

Step-by-step explanation:

The geometric distribution is the number of trials required to have r successes. The measures the number of sucesses(wins), not the number of trials required to win r games. So the geometric distribution does not apply.

For each match, there are only two possible outcomes, either the skilled player wins, or he does not. The probability of the skilled player winning a game is independent of other games. So the binomial distribution applies.

We can also find the expected number of wins of the skilled player, which is 15*0.9 = 13.5. The Poisson distribution is a discrete distribution in which the only parameter is the expected number of sucesses. So the Poisson distribution applies.

So the correct answer is:

b) Binomial

c) Poisson

7 0

3 years ago

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Assuming the incident of fires for individual reactors can be described by a Poisson distribution, what is the probability that

Maksim231197 [3]

Complete Question

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

If b is between a and c then ab+bc=ac

erma4kov [3.2K]

That's false. Take the numbers 1, 3, and 5.
Let a = 1, b = 3, and c = 5.
ab + bc = ac
3 + 15 = 5
18 ≠ 5
Since 18 does not equal 5, this is false. If you were to use the distributive property on ab + bc = ac, you would get
b(a+c) = ac, which doesn't even make sense.
Have a nice day! :)

3 0

3 years ago