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The answers are ONM, NOM, and alternate interior angles.
Based on the SSS postulate, the two triangles △MLO and △ONM are congruent since three sides of △MLO are respectively equal to the three sides of △ONM.
Based on CPCTC, all of the corresponding angles of △MLO and △ONM are congruent as well since the two triangles are congruent, that is,
∠LMO≅∠NOM and ∠NMO≅∠LOM.
Since the pair ∠LMO and ∠NOM as well as ∠NMO and ∠LOM are angles on the inner side of two lines but on opposite sides of the transversal MO, these pairs of angles are also alternate interior angles.

5 0

3 years ago

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Two corporate baseball teams are scheduled to play a game together. They agree that if both teams attend or if neither team atte

sp2606 [1]

Answer:

2.99% probability that the cost will be paid by only one team

Step-by-step explanation:

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.

Probability that a team plays:

A team plays if it has at most 2 injured players out of 11.

11 players, so $n = 11$

Each player with a 5% probability of injury, so $p = 0.05$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

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

$P(X = 0) = C_{11,0}.(0.05)^{0}.(0.95)^{11} = 0.5688$

$P(X = 1) = C_{11,1}.(0.05)^{1}.(0.95)^{10} = 0.3293$

$P(X = 2) = C_{11,2}.(0.05)^{2}.(0.95)^{9} = 0.0867$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5688 + 0.3293 + 0.0867 = 0.9848$

Each team has a 0.9848 probability of showing up to play.

What is the probability that the cost will be paid by only one team?

This happens if one team shows up and the other do not.

2 teams, so $n = 2$

Each team has a 0.9848 probability of showing up to play, so $p = 0.9848$ .

This probability is P(X = 1).

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

$P(X = 1) = C_{2,1}.(0.9848)^{1}.(0.0152)^{1} = 0.0299$

2.99% probability that the cost will be paid by only one team

7 0

4 years ago

If DE = 37 cm and EF = 16 cm, then what are the possible lengths for DF so that DE,EF, and DF can form a triangle? Explain your

Molodets [167]

Answer:

The length of DF must be between 21 and 53.

Step-by-step explanation:

In a triangle, the length of two sides added together must exceed the length of the 3rd side. So, since EF is the shortest of the two givens, we know that EF + DF must be greater than DE. So we can plug in these numbers to find the minimum.

EF + DF > DE

16 + DF > 37

DF > 21

Now, for the upper maximum, we know that the two given lengths must be greater than the length of DF. So again, we can solve for the maximum using the amounts.

DE + EF > DF

37 + 16 > DF

53 > DF

With these two in mind, we know that DF must be between 21 and 53

6 0

4 years ago