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2 years ago

In order for a rhombus to be proven using either its sides or angels, it must first be proven to be a

Mathematics

1 answer:

Paladinen [302]2 years ago

5 0

Answer:

It must be proven to be a quadrilateral

Step-by-step explanation:

The question is straightforward, and it requires a direct answer.

A rhombus has four sides and as such it means it is a type of quadrilateral.

So, before a shape is proven to be a rhombus, it has to first be proven to be a quadrilateral.

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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

2 years ago

8. Is the following statement true or false? Justify your conclusion. If a and b are nonnegative real numbers and a + b =0, then

atroni [7]

Answer:

If $a + b = 0$ then $a = 0$ and $b =0$

a | b | a + b (answer)

0 | 0 | 0

0 | 1 | 1

0 | 2 | 2

1 | 0 | 1

2 | 0 | 2

1 | 1 | 2

2 | 1 | 3

Step-by-step explanation:

Considering the following conditions for the real numbers:

$a\geq 0\\b\geq 0$

Following the rules of these in-equations, it is possible to deduce:

$a + b \geq 0$

Then, if the proposed statement is:

$a + b = 0$

The conditions above shall comply the requirements established, but first, analyzing the statement:

If $a \geq 0$ and $b \geq 0$ then $a + b \geq a$ , $a + b \geq b$ and $a + b \geq 0$ .

If $a = 0$ and b a non negative real number, then $a + b \geq b$ , but because to $a = 0$ , then $a + b = b$ . Due to the commutative property of sums, the same behavior will be presented if $b = 0$ and a a non negative real number.

According to that, if $a + b = 0$ , then $a = 0$ and $b = 0$ .

7 0

3 years ago

Find the product. (y^3)^2 * y^7

Natasha_Volkova [10]

Firstly, let's take first factor:
(y^3)^2 = y^(3*2) = y^6
and then:
y^6 * y^7 = y^(6 + 7) = y^13

8 0

2 years ago

I need this done today whoever gets this correct gets brainliest!!! PLZZ HELPP MEE!!!

Stella [2.4K]

Answer:

B Survey a random sample of eighth-grade students

Step-by-step explanation:

In order to reach the best conclusion without biased data, the best sampling method would be B because not only does it survey 8th graders, but it also surveys a random group, ensuring that not all of the students are potentially excelling in and/or having trouble with homework.

7 0

2 years ago

Hereeeeeee helllllllpp lol

Ivenika [448]

It should be 0 because the only factor changing is the x value

7 0

2 years ago