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blagie [28]
3 years ago
7

Answer asap! will give brainliest! question is in pic

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

Answer:

UW=14

Step-by-step explanation:

First we know that UV+VW=UW, so:

4+x+10=2x+14

Solve for x first:

x+10=2x+10

10=x+10

x=0

Then, substitute the X

2(0+14)=14=UW

To check your answer, substitute and add UV and VW

4+(0)+10

14

<em>Hope this helped! Have a good day!</em>

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What is the volume of a sphere with a radius of 4
Ede4ka [16]

Step-by-step explanation: To find the volume of a sphere, start with the formula for the volume of a sphere which is shown below.

V = \frac{4}{3}\pi  r^{3}

Here, we are given that our sphere has a radius of 4 units.

So plugging into the formula, we have (\frac{4}{3})(\pi)(4 units)^{3}.

Start by simplifying the exponent.

(4 units)³ is equal to (4 units) (4 units) (4 units) or 64 units³.

So we have (\frac{4}{3})(\pi)(64 units^3}).

Next, we multiply (4/3)(64) which can

be thought of as (4/3)(64/1)

So multiplying across the numerators and across the denominators,

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5 0
3 years ago
I just need examples of these things, I have to make my own problems and solve them:
poizon [28]
1) x+17=20
x-10=30

2) x/4=20
x/30=2

3) 4(x+5)=40

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Hope this helps! Have a wonderful day.
6 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!}&#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}

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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
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Answer:

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Schach [20]
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