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sveticcg [70]
3 years ago
8

Help sos i don’t know how to do this

Mathematics
1 answer:
nekit [7.7K]3 years ago
6 0

9514 1404 393

Answer:

  (a)  2

Step-by-step explanation:

Simplify this fraction the same way you would any fraction: cancel common factors from numerator and denominator.

  22x^2y^2\div11x^2y^2=\dfrac{22x^2y^2}{11x^2y^2}=\dfrac{2\cdot11x^2y^2}{11x^2y^2}=2\cdot\dfrac{11x^2y^2}{11x^2y^2} = 2\cdot1 = \boxed{2}

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Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1
alexandr1967 [171]

Answer:

a) \simeq 0.3012   b) \simeq 0.0494 c) \simeq 0.2438

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, \frac{1.2}{4}

= 0.3 collisions per  month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

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


                                                           for x ∈ N ∪ {0}

                       =  0 otherwise --------------------------------------(1)

here, \lambda = 0.3 collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

         P(X = 0) = \frac{e^{-0.3}\times {\0.3}^{0}}{0!}


                      \simeq 0.7408182207 ------------------------------------(2)

Now, since we are calculating  this for 4 months,

so, P(No collision in 4 month period)

     =0.7408182207^{4}

     \simeq 0.3012  -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

           P(X =1) = \frac{e^{-0.3}\times {\0.3}^{1}}{1!}


                      \simeq 0.2222454662 ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

                =0.2222454662^{2}

                \simeq 0.0494 -----------------------------------------(5)

1 collision in 6 months period means

                                \frac{1}{6} collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6]  (which is to be estimated)

=\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

\simeq 0.6543894283-------------------------------------------(6)

So,

P(1  collision in 6 month period)

  =  0.6543894283^{6}

   \simeq 0.0785267444 ------------------------------------------------(7)

So,

P(No collision in 6 months period)

  = (P(X =0)^{6}

   \simeq 0.1652988882 ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

\simeq  0.2438 ---------------------------------------------(9)          

7 0
3 years ago
Please help meeeeeee!!!!!
Tasya [4]

Answer: A

Step-by-step explanation:

This is because 25%= 25/100, and that reduces to 1/4. Also, whenever it says something OF something, it means to multiply. In this case, 1/4 OF 1000= 1/4x1000= 250.

*Even though it seems that multiplying will just make the number bigger, when you multiply with a fraction, it actually gets smaller* For example 1/3 of 3= 1/3x3= 1

8 0
2 years ago
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John had 2/3 of his homework complete, Sarah had 5/10, Alex 7/8, and Michelle 1/2. Who competed the most of the homework assignm
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Alex definitely had the most work done between all of them
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How do you get that m
Otrada [13]

Answer:

What?

Step-by-step explanation:

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2 years ago
HELP MUST BE DONE BY 4:00
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Answer:

the antelope is 5280 feet the cougar is 4224 feet the hare is 4136 feet the kangaroo is 3520 feet and the coyote is 3773 feet the ostrich is 3773 feet hope this helps the<em> answer is the </em><em>antelope</em>

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