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Mrrafil [7]
3 years ago
6

Will you please help me solve this ​

Mathematics
1 answer:
likoan [24]3 years ago
4 0

Answer:

-18

Step-by-step explanation:

Substitute in -2 for x

-(-2)²+2(-2)-10

-4-4-10

-18

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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
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kirza4 [7]
Additive inverses combine to get 0. So, k is -1.4. The sum is 0. Hopefully that will help.
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Please help me. I have no idea what to do!!
boyakko [2]

Answer:

It's the first option.

Step-by-step explanation:

It is turned down so the one half is negative (-1/2)

parabola will be in the form of a(x-h)^2 + k. (-h, k) is your vertex, in this case that is (15,25). so in the equation, you will have

-1/2(x-15)^2 + 25. first one.

Hope this helps!

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Mazyrski [523]

Answer: a and c

Step-by-step explanation: both have equal hight

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3 years ago
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mestny [16]
Basically, we need to plug in the given values for the variables x and y into the given expression 1.3(x+2)-(y-6). First off, we can plug in all the given values into the expression, giving us 1.3(4+2)-(2-6). Now, perform the operations on the inside of the parentheses. Doing this, we get 1.3(6)-(-4). Now, we use the distributive property to simplify. This gives us 7.8+4. Finally, when we add the two numbers, we get \boxed{11.8}. Hope this helped!
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