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BaLLatris [955]
3 years ago
13

Which number do you add to 3 to get 0?

Mathematics
2 answers:
julia-pushkina [17]3 years ago
8 0

Answer:

negative 3, (-3)

Step-by-step explanation:

grandymaker [24]3 years ago
7 0

Answer:

-3

Step-by-step explanation:

3 + (-3) = 0

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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
Please help me out! And show your work :)) <3
8_murik_8 [283]

Answer:

mbnmf mkvdv dlbj fbljkf mgmbv m

Step-by-step explanation:

vk gmb flkmvfvv

3 0
3 years ago
Read 2 more answers
What is the mass of 20 gallons of gasoline?
Tomtit [17]
Your answer would be 14.978
3 0
3 years ago
Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$
Viefleur [7K]

QUESTION 1

Given that:

f(x)=2x^2+3x-9,

g(x)=5x+11,

and

h(x)=-3x^2+1

Then;

f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)

f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1

Group similar terms;

f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1

Simplify;

f(x)-g(x)+h(x)=-x^2-2x-19

QUESTION 2

Given that;

f(x)=4x-7.

g(x)=(x+1)^2

and

s(x)=f(x)+g(x)

Substitute the functions;

s(x)=4x-7+(x+1)^2

Substitute x=3

s(3)=4(3)-7+(3+1)^2

s(3)=12-7+(4)^2

s(3)=5+16

s(3)=21

QUESTION 3

Given:

f(x)=3x+2

g(x)=x^2-5x-1

f(g(x))=f(x^2-5x-1)

This implies that;

f(g(x))=3(x^2-5x-1)+2

Expand the parenthesis;

f(g(x))=3x^2-15x-3+2

f(g(x))=3x^2-15x-1

QUESTION 4

The given function is;

f(x)=3(x-6)^2+1

Let

y=3(x-6)^2+1

\Rightarrow y-1=3(x-6)^2

\Rightarrow \frac{y-1}{3}=(x-6)^2

\Rightarrow \sqrt{\frac{y-1}{3}}=x-6

\Rightarrow x=6+\sqrt{\frac{y-1}{3}}

The range is:

\frac{y-1}{3}\ge0

y-1\ge0

y\ge1

The interval notation is;

[1,+\infty)

6 0
3 years ago
Alicia needs to waterproof her house’s back deck, which is 18 feet by 5 feet. The waterproof coating costs $6 for a can that cov
const2013 [10]

Step-by-step explanation:

18 × 5 = 90

90%30=3

3×6=18

4 0
2 years ago
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