All categories

2 years ago

Parallel Lines and Paris of Angles

Mathematics

2 answers:

Sedaia [141]2 years ago

8 0

Answer:

The explanation on how to do them is down below. If you have anymore questions pls feel free to ask

Step-by-step explanation:

So for number one you just need to state the sentence under the lines back wards but starting Lines m and n......

For number two name two angles that are inside the parallel lines and for three the opposite name two that are basically verticle outside of the parallel lines.

For number four they are across from each so they are verticle and since they are verticle they are concurrent so the m of 2 is 125.

In number 5,4 and 7 are corresponding so the are equal.

In number 6 the angles are alternate interior so they are congruent.

In number 7, the angles are corresponding so they are congruent.

In number 8, the angles are alternate exterior so they are congruent.

In number 9, the angles are are verticle so the are congruent.

In number 10, the answer is supplementary because on the top and bottom they add up to 180 degrees. And it part b 2 and 3 are adjacent and supplementary, so you subtract 119 from 180 and you get 61.

tresset_1 [31]2 years ago

7 0

I’m not sure but I think number 1 is m and n

You might be interested in

United Flight 15 from New York's JFK airport to San Francisco uses a Boeing 757-200 with 182 seats. Because some people with res

Tcecarenko [31]

Answer:

There is a 29.27% probability that the flight is overbooked. This is not an unusually low probability. So it does seem too high so that changes must be made to make it lower.

Step-by-step explanation:

For each passenger, there are only two outcomes possible. Either they show up for the flight, or they do not show up. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

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

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

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

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

And $\pi$ is the probability of X happening.

A probability is said to be unusually low if it is lower than 5%.

For this problem, we have that:

There are 200 reservations, so $n = 200$ .

A passenger consists in a passenger not showing up. There is a .0995 probability that a passenger with a reservation will not show up for the flight. So $\pi = 0.0995$ .

Find the probability that when 200 reservations are accepted for United Flight 15, there are more passengers showing up than there are seats available.

X is the number of passengers that do not show up. It needs to be at least 18 for the flight not being overbooked. So we want to find $P(X < 18)$ , with $\pi = 0.0995, n = 200$ . We can use a binomial probability calculator, and we find that:

$P(X < 18) = 0.2927$ .

There is a 29.27% probability that the flight is overbooked. This is not an unusually low probability. So it does seem too high so that changes must be made to make it lower.

5 0

3 years ago

G+3>6<br> A (3)<br> B (4)<br> C (-3)<br> D (-4)

aleksley [76]

G>6-3
G>3 is definitely the answer

7 0

3 years ago

If i flip a fair coin six times, what is the probability that three of the flips lands on heads?

aalyn [17]

3/6 which can be 50 percent chances of getting!

8 0

4 years ago