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Airida [17]
3 years ago
9

Adding which terms to 3x2y would result in a monomial? Check all that apply.

Mathematics
2 answers:
kkurt [141]3 years ago
8 0
 the answers are -12x^2 and 4x^2y. hopefully it helps and i just got it right.
velikii [3]3 years ago
5 0

Answer:

option 2 and 6

Step-by-step explanation:

Given : Monomial 3x^2y

To find : Adding which terms to  3x^2y would result in a monomial? Check all that apply.

Solution :

Monomial is when the number of terms is only 1.

1) 3xy

Adding in given monomial

3x^2y+3xy     is a binomial

2) -12x^2y

Adding in given monomial

3x^2y+(-12x^2y)=-9x^2y     is a monomial

3) 2x^2y^2

Adding in given monomial

3x^2y+2x^2y^2     is a binomial

4) 7xy^2

Adding in given monomial

3x^2y+7xy^2     is a binomial

5) -10x^2

Adding in given monomial

3x^2y-10x^2     is a binomial

6) 4x^2y

Adding in given monomial

3x^2y+4x^2y=7x^2y is a monomial

7) 3x^3

Adding in given monomial

3x^2y+3x^3     is a binomial

Therefore, Option 2 and 6 are the terms giving a monomial.

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Suppose that minor errors occur on a computer in a space station, which will require re-calculation. Assume the occurrence of er
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Answer:

a

  P(X = 0) =  0.6065

b

P(x <  25 ) =   1.18 *10^{-33}

c

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Step-by-step explanation:

From the question we are told that

  The rate  is \lambda =  \frac{1}{2}\   hr^{-1}    =  0.5 / hr

  Generally  Poisson distribution formula is mathematically represented as

       P(X = x) =  \frac{(\lambda t) ^x e^{-\lambda t }}{x!}

Generally the probability that no error occurred during a day is mathematically represented as  

Here  t =  1  hour according to question a

So

   P(X = x) =  \frac{\lambda^x e^{-\lambda}}{x!}

Hence

   [tex]P(X = 0) =  \frac{\frac{1}{2} ^0 e^{-\frac{1}{2}}}{0!}

=>  P(X = 0) =  0.6065

Generally the probability that  a critical error occurs since the start of a day is mathematically represented as

Here  t =  1  hour according to question a

So

   P(X = x) =  \frac{\lambda^x e^{-\lambda}}{x!}

Hence

      P(x \ge 25 ) =  1 - P(x <  25 )

Here

     P(x <  25 ) = \sum_{x=0}^{24} \frac{e^{-\lambda} * \lambda^{x}}{x!}

=>   P(x <  25 ) =  \frac{e^{-0.5} *0.5^{0}}{0!} + \cdots + \frac{e^{-0.5} *0.5^{24}}{24!}

P(x <  25 ) =  0.6065 + \cdots + \frac{e^{-0.5} *0.5^{24}}{6.204484 * 10^{23}}

P(x <  25 ) =  0.6065 + \cdots + 6.0*10^{-32}

P(x <  25 ) =   1.18 *10^{-33}

Considering question c

Here  t =  2  

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=> P(x \le 5 ) =   \frac{(0.5 *  2) ^ 0 e^{- 0.5  * 2 }}{0!} + \cdots  +   \frac{(0.5 *  2) ^ 5  e^{- 0.5  * 2 }}{5!}

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=>  P(x \le 5 ) =   2.7183 + \cdots  +   0.0226525

 P(x \le 5 ) =  0.9994    

8 0
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