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Xelga [282]
3 years ago
9

Give an example of parallel lines in the real world. Then describe how you could prove that the lines are parallel.

Mathematics
1 answer:
zhuklara [117]3 years ago
4 0
The sides of a TV.

since left and right (sides) goes up and down


---------------
| T.V | <-- SIDES R THE SAME VERTICAL
| °°° | LINE <UP AND DOWN>
---------------
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−\frac{7}{10} ÷\frac{2}{5}
bazaltina [42]
$$ −\frac{7}{10} \div\frac{2}{5}= \frac{-7}{4} $$ . totally answer. I hope helping with this answer
3 0
3 years ago
(15, 31) is rotated 90 ° counterclockwise which coordinates gives me
dezoksy [38]

Answer:

(-31,15)

Step-by-step explanation:

Rotating 90 degrees clockwise is very simple:

(a,b) 90 degrees= (-b,a)

Very simple.

Just switch sides and add negative to the first term.

4 0
4 years ago
b) How many ways can you deal cards (from a deck of 52) to 4 people when each player gets 7 cards. 2 hidden and 5 visible. Assum
Mariana [72]

Answer:

{52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

Step-by-step explanation:

The dealing of the cards can be seen in the following way:

We first have to choose which 7 cards are going to be dealt to the 1st player. So we have to pick 7 cards out of the 52 available cards. This can be done in {52 \choose 7} ways. Now that we have chosen which 7 cards are going to be dealt to the 1st player, we have to choose which 2 of them are going to be the hidden ones. So we have to pick 2 cards out of the 7 cards to be the hidden ones. This can be done in {7 \choose 2} ways. At this point we now know which cards are being dealt to the 1st player, and which ones are hidden for him. Then we have to choose which 7 cards to deal to the 2nd player, out of the remaining ones. So we have to pick 7 out of 52-7=45. (since 7 have been already dealt to the 1st player). This can be done in {45 \choose 7}. Then again, we have to pick which 2 are going to be the hidden cards for this 2nd player. So we have to pick 2 out of the 7. This can be done in {7 \choose 2} ways. Then we continue with the 3rd player. We have to choose 7 cards out of the remaining ones, which at this point are 52-7-7=38. This can be done in {38 \choose 7}. And again, we have to choose which ones are the hidden ones, which can be chosen in {7 \choose 2} ways. Finally, for the last player, we choose 7 out of the remaining cards, which are 52-7-7-7=31. This can be done in {31 \choose 7} ways. And choosing which ones are the hidden ones for this player can be done in {7 \choose 2} ways. At the end, we should multiply all our available choices on each step, to get the total choices or total ways to deal the cards to our 4 players (since dealing the cards is a process of several steps with many choices on each step).

{52 \choose 7}{7 \choose 2}{45 \choose 7}{7 \choose 2}{38 \choose 7}{7 \choose 2}{31 \choose 7}{7 \choose 2}={52 \choose 7}{45 \choose 7}{38 \choose 7}{31 \choose 7}{7 \choose 2}^4

7 0
3 years ago
PLEASE HELP ASAP!!! :)) ITS EASY
inn [45]

Answer:

B

Step-by-step explanation:

B

8 0
3 years ago
Read 2 more answers
A real estate agent has 19 properties that she shows. She feels that there is a 30% chance of selling any one property during a
netineya [11]

Answer:

P(X \geq 5)=1-P(X

We can find the individual probabilities:

P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114

P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092

P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358

P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869

P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491

And replacing we got:

P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=19, p=0.3)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

And we want to find this probability:

P(X \geq 5)

And we can use the complement rule:

P(X \geq 5)=1-P(X

We can find the individual probabilities:

P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114

P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092

P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358

P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869

P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491

And replacing we got:

P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178

4 0
3 years ago
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