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2 years ago

Which pairs of angles in the figure below are vertical angles?

Mathematics

1 answer:

andre [41]2 years ago

5 0

Answer:

C & D because vertical angles have the same measures and thoses angles intersects each other

Step-by-step explanation:

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HELP! Find the Area of the triangle

Zigmanuir [339]

Answer:

A ≈ 65.8 m²

Step-by-step explanation:

Given 2 sides and the angle between them then the area (A) is

A = $\frac{1}{2}$ bc sinA

= 0.5 × 19.9 × 10.7 × sin38.17°

≈ 65.8 m² ( to 1 dec. place )

6 0

2 years ago

Find the coordinates of the centroid of triangle RST. SHOW YOUR WORK so I can see if the answer makes sense (no links)

zimovet [89]

Note the coordinates of each point: R(-4, 5), S(5, 1), T(2, -3).

The centroid is the point whose coordinates are the average of the coordinates of R, S, and T.

x-coordinate: (-4 + 5 + 2)/3 = 3/3 = 1

y-coordinate: (5 + 1 - 3)/3 = 3/3 = 1

So the centroid is (1, 1).

3 0

2 years ago

Question 15 (5 points)

serg [7]

Answer:

You need to attach the pictures and ask again.

Step-by-step explanation:

I cannot see any attachments.

8 0

2 years ago

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$