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lina2011 [118]
2 years ago
5

Based on these segment lengths, which group of segments cannot form a triangle? a. 12, 7, 8 b. 8, 7, 13 c. 1, 2, 3 d. 80, 140, 7

0
Mathematics
1 answer:
TEA [102]2 years ago
5 0

(D) 80°, 140°, and 70° group of segments cannot form a triangle.

<h3>What is a triangle?</h3>
  • A triangle is a three-edged polygon with three vertices.
  • It is a fundamental shape in geometry.
  • Triangle ABC represents a triangle with vertices A, B, and C.
  • In Euclidean geometry, any three non-collinear points define a unique triangle and, by extension, a unique plane.
  • In other words, the triangle is contained in just one plane, and every triangle is contained in some plane.
  • There is just one plane and all triangles are enclosed in it if the entire geometry is merely the Euclidean plane; but, in higher-dimensional Euclidean spaces, this is no longer true.

To find which group of segments cannot form a triangle:

  • 80°, 140°, and 70° cannot form a triangle because the sum of the three angles is 290°, whereas the sum of the angles in a triangle is 180°.

Therefore, (D) 80°, 140°, and 70° group of segments cannot form a triangle.

Know more about triangles here:

brainly.com/question/17335144

#SPJ4

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notsponge [240]

Answer:

0.8895 = 88.95% probability that the hockey team wins at least 3 games in November.

Step-by-step explanation:

For each game, there are only two possible outcomes. Either the teams wins, or they do not win. The probability of the team winning a game is independent of any other game, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The probability that a certain hockey team will win any given game is 0.3773.

This means that p = 0.3773

Their schedule for November contains 12 games.

This means that n = 12

Find the probability that the hockey team wins at least 3 games in November.

This is:

P(X \geq 3) = 1 - P(X < 3)

In which:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{12,0}.(0.3773)^{0}.(0.6227)^{12} = 0.0034

P(X = 1) = C_{12,1}.(0.3773)^{1}.(0.6227)^{11} = 0.0247

P(X = 2) = C_{12,2}.(0.3773)^{2}.(0.6227)^{10} = 0.0824

Then

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0034 + 0.0247 + 0.0824 = 0.1105

P(X \geq 3) = 1 - P(X < 3) = 1 - 0.1105 = 0.8895

0.8895 = 88.95% probability that the hockey team wins at least 3 games in November.

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