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Oksanka [162]
2 years ago
3

Points A, B, C, and D are vertices of a square. Point E is inside the square. Explain how to tell whether point E is closer to A

, B, C, or D.
Mathematics
1 answer:
arlik [135]2 years ago
4 0

Answer:

To determine if a point E inside a square is closer to any of the vertices of the square we have to determine the length of the diagonals between the point E and the vertices of the square i.e. AE, BE, CE, DE

The diagonal can be determined by applying Pythagorean's theorem

Step-by-step explanation:

To determine if a point E inside a square is closer to any of the vertices of the square we have to determine the length of the diagonals between the point E and the vertices of the square i.e. AE, BE, CE, DE

The diagonal can be determined by applying Pythagorean's theorem

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You and a friend each randomly draw a card from a standard deck. What is the probability that at least one of you is holding a f
Andreyy89

Answer:

0.24

Step-by-step explanation:

Total cards = 52

Face cards = heart {K,Q,J} , Spade {K,Q,J} , Club {K,Q,J} , Diamond{K,Q,J} = 12

Now we are given that You and a friend each randomly draw a card from a standard deck.

Probability that no one getting face card:

Favorable events = total cards - face cards = 52 -12= 40

Total cards = 52

So, probability that no one is getting face card = \frac{40}{52} =0.76

Now we are supposed to find the probability that at least one of you is holding a face card

So, probability = 1 - probability that no one is getting face card

                        =1 - 0.76

                        = 0.24

So, probability that at least one of you is holding a face card is 0.24

8 0
2 years ago
Read 2 more answers
Analyze the diagram to complete the statements. The m∠MXN is the m∠YZX. The m∠LZX is the m∠ZYX + m∠YXZ. The m∠MYL is 180° − m∠ZY
Xelga [282]

By using the picture that was provided below, you can see that

m<MXN is greater than m<YZX.  

Angles m<LZX is equal to m<ZYX + m<YZX and

m<MYL is equal to 180 degrees minus M<ZYX.


I hope this helps! Have a good day


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2 years ago
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The function y = (x + 4)4 is a transformation of the graph of the parent function y = x4. How is the zero of the parent function
Musya8 [376]
The answer is C but not sure about it please double check it


Thank you

I hope it’s help you
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2 years ago
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Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang
kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145

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