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Leona [35]
3 years ago
12

In the figure given alongside, angle a=angle x and angle b= angle y, show that angle x + angle y + angle z= 180°​

Mathematics
2 answers:
alexandr402 [8]3 years ago
8 0

Step-by-step explanation:

x+y+z= 180 (being sum of three angle of a triangle)

kykrilka [37]3 years ago
5 0

Here,

Given,

<a = <x and <b = <y

To prove = <x+ <y + <z= 180°

Then,

<a + <b + <z = 180° ( St. angle )

▲ <x + <y + <z = 180°

Hence, Proved

I hope you understand...

Thanks...

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Step-by-step explanation:

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What is the answer to the equation 2x = -10
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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!}&#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
3 years ago
Helppppppppppppppppppppppppppppppppppppp
sineoko [7]

Answer:

Option C is correct.........

4 0
3 years ago
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