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Umnica [9.8K]
3 years ago
12

In an experiment, a fair coin is tossed 13 times and the face that appears (H for head or T for tail) for each toss is recorded.

Mathematics
1 answer:
zalisa [80]3 years ago
5 0

Answer:

1) 1 element

2) 13 elements

3) 22 elements

4) 40 elements

Step-by-step explanation:

1) Only one element will have no tails: the event that all the coins are heads.

2) 13 elements will have exactly one tile. Basically you have one element in each position that you can put a tail in.

3) There are {13 \choose 2} = 78 elements that have exactly 2 tails. From those elements we have to remove the only element that starts and ends with a tail and in the middle it has heads only and the elements that starts and ends with a head and in the 11 remaining coins there are exactly 2 tails. For the last case, there are {11 \choose 2} = 55 possibilities, thus, the total amount of elements with one tile in the border and another one in the middle is 78-55-1 = 22

4) We can have:

  • A pair at the start/end and another tail in the middle (this includes a triple at the start/end)
  • One tail at the start/end and a pair in the middle (with heads next to the tail at the start/end)

For the first possibility there are 2 * 11 = 22 possibilities (first decide if the pair starts or ends and then select the remaining tail)

For the second possibility, we have 2*9 = 18 possibilities (first, select if there is a tail at the end or at the start, then put a head next to it and on the other extreme, for the remaining 10 coins, there are 9 possibilities to select 2 cosecutive ones to be tails).

This gives us a total of 18+22 = 40 possibilities.

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51.2 h = 839.68

h = 839.68 / 51.2

h = 16.4

So, your answer is 16.4

Hope this helps!
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I sphere has a diameter of 24 units what is the volume in cubic units
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4π(r³/3)=V
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What % is 600,000 of 65 million
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Rewrite 7 − 8 using the additive inverse and display the new expression on a number line.
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4 0
11 months ago
Problem: Match each value on the left with the number on the right that includes that value.
Anestetic [448]

Answer:

5 Thousands ⇒ 45,089

8 Hundreds ⇒ 958,823

3 millions ⇒ 3,072,700

2 Ten thousands ⇒ 7,426,580

7 Hundred Thousands ⇒ 752,671

Step-by-step explanation:

Solution:

We need match the numbers with given set of numbers.

Now,

5 thousands ⇒

So we can say that from given set of the numbers we need to find the number 5 belongs in thousands place.

So we can say that the number 5 in thousands place from given of number is ;

5 Thousands ⇒ 45,089

8 Hundreds ⇒

So we can say that from given set of the numbers we need to find the number 8 belongs in hundred place.

So we can say that the number 8 in Hundred place from given of number is ;

8 Hundreds ⇒ 958,823

3 Millions ⇒

So we can say that from given set of the numbers we need to find the number 3 belongs in millions place.

So we can say that the number 3 in Millions place from given of number is ;

3 millions ⇒ 3,072,700

2 Ten thousands ⇒

So we can say that from given set of the numbers we need to find the number 2 belongs in Ten thousands place.

So we can say that the number 3 in Ten Thousands place from given of number is ;

2 Ten thousands ⇒ 7,426,580

7 Hundred thousands ⇒

So we can say that from given set of the numbers we need to find the number 7 belongs in Hundred thousands place.

So we can say that the number 7 in Hundred Thousands place from given of number is ;

7 Hundred Thousands ⇒ 752,671

Hence the matching numbers are

5 Thousands ⇒ 45,089

8 Hundreds ⇒ 958,823

3 millions ⇒ 3,072,700

2 Ten thousands ⇒ 7,426,580

7 Hundred Thousands ⇒ 752,671

6 0
2 years ago
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