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taurus [48]
3 years ago
5

PLEASE HELP

Mathematics
2 answers:
elixir [45]3 years ago
5 0

Answer:

AB = (-1–3)/2 = -2

AC: (7+1)/6 = 4/3

BC: (7–3)/(6+2) = 1/2

Step-by-step explanation:

Galina-37 [17]3 years ago
3 0

Answer:

Step-by-step explanation:

34.546.567

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he production of pipes has a mean diameter of 3.25 inches and a standard deviation of .15 inches. The shape of the distribution
o-na [289]

Answer:

The probability that a randomly chosen part has diameter of 3.5 inches or more is 0.9525

Step-by-step explanation:

Mean = \mu = 3.25 inches

Standard deviation = \sigma = 0.15 inches

We are supposed to determine the probability that a randomly chosen part has diameter of 3.5 inches or more

P(Z \geq 3.5)=1-P(z

Refer the z table for p value

P(Z \geq 3.5)=1-0.0475\\P(Z \geq 3.5)=0.9525

Hence  the probability that a randomly chosen part has diameter of 3.5 inches or more is 0.9525

7 0
3 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!}


                                                           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
OK right image and easier to read lol
Brums [2.3K]

Answer:

you want us to do all that

Step-by-step explanation:

6 0
3 years ago
How are you supposed solve x
Sergeeva-Olga [200]
Because the lines are parallel, 75 is equal to 11x-2
75=11x-2
77=11x
x=7
7 0
3 years ago
Read 2 more answers
Please help meee and show all steps please!!!ty
kolbaska11 [484]

Answer:

x = -4 and x = 5

Step-by-step explanation:

Since x^2 + 2x and 3x + 20 both equal to y, we know that the expressions equal to each other. We can write a new equation base on that.

x^2 + 2x = 3x + 20

Now we solve the equation.

x^2 + 2x = 3x + 20

x^2 - x = 20

x^2 - x - 20 = 0

(x + 4) (x - 5) = 0

x + 4 = 0 ; x -5 = 0

x = -4 ; x = 5

3 0
2 years ago
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