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Sonja [21]
3 years ago
9

The concurrency of which lines create the Centroid?

Mathematics
1 answer:
meriva3 years ago
4 0
The answer for that question is Angle bisectors
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Find the length of the unknown side. Round your answer to the nearest tenth.
slamgirl [31]

Answer:

11.5

Step-by-step explanation

a^2+b^2=c^2

8(8)+(x)^2=196

64 + x^2=196

\sqrt{x^{2} }=\sqrt{132}

x=11.48

x=11.5 rounded

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3 years ago
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Complete this statement:
klasskru [66]

Answer:

45x^3a+27xa^2= 9xa(5x^2+3a)

Step-by-step explanation:

Given:

45x^3a+27xa^2= 9xa

We need to complete this Statement.

By Solving the above equation we get;

We will take some common factor out so we will get;

45x^3a+27xa^2= 9xa(5x^2+3a)

Hence the Complete statement is 45x^3a+27xa^2= 9xa(5x^2+3a)

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2 years ago
Increase 40 by 1/2 <br> I need help, im bad at mathematics.
Bess [88]

Answer:

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Step-by-step explanation:

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6 0
2 years ago
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!}&#10;

                                                           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!}&#10;

                      \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!}&#10;

                      \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
2 years ago
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