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3 years ago

Note: Figure is not drawn to scale.

Mathematics

1 answer:

marishachu [46]3 years ago

6 0

The correct answer is A. 124°

Further explanation:

First of all we will have to define alternate interior angles.

"When two parallel lines are intersected by a traversal, the angles which are formed inside the parallel lines are called alternate interior angles"

In the given figure,

The alternate angles are:

m∠O and m∠X

m∠W and m∠P

We are given that m∠W is 124°.

The alternate angles' theorem states that the alternate angles are equal.

So through the theorem,

m∠W = m∠P = 124°

The correct answer is A. 124°

Keywords: Alternate angles, Parallel Lines

Learn more about parallel lines at:

#LearnwithBrainly

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Anton [14]

Answer:

33.94 cms of ribbon

Step-by-step explanation:

Because you need two diagonals to form the x, therefore the amount of ribbon needed is the sum of the distance of both diagonals.

When crossing the diagonal, a rectangular angle is formed, where the diagonal would be the hypotenuse, we know that the distance of the hypotenuse can be calculated by means of the legs, which we know its value (12):

d ^ 2 = a ^ 2 + b ^ 2

a = b = 12

d ^ 2 = 12 ^ 2 + 12 ^ 2

d ^ 2 = 288

d = 288 ^ (1/2)

d = 16.97

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Answer from jmonterrozar

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2 years ago

What is the slope of the line that contains the point -6,-7 and -6,2

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We know that coordinates are labeled (x,y)
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3 years ago

Read 2 more answers

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,

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= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

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Step-by-step explanation:

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