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kakasveta [241]
3 years ago
5

In a race in which eleven automobiles are entered and there are no ties, in how many ways can the first three finishers come in?

Mathematics
1 answer:
viva [34]3 years ago
8 0

Answer:

990 ways

Step-by-step explanation:

The total number of automobiles we have is 11.

Now, what this means is that for the first position , we shall be selecting 1 out of 11 automobiles, this can be done in 11 ways( 11C1 = 11!/(11-1)!1! = 11!/10!1! = 11 ways)

For the second position, since we have the first position already, the number of ways we can select the second position is selecting 1 out of available 10 and that can be done in 10 ways(10C1 ways = 10!9!1! = 10 ways)

For the third position, we have 9 automobiles and we want to select 1, this can be done in 9 ways(9C1 ways = 9!/8!1! = 9 ways)

Thus, the total number of ways the first three finishers come in = 11 * 10 * 9 = 990 ways

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vesna_86 [32]

The first two numbers between 100 and 150 that have a HCF of 22 are 110 and 132

<h3>Highest common factors</h3>

The greatest common divisor of two or more integers, which are not all zero, is the largest positive integer that divides each of the integers.

According to the question, we are to find two numbers between 100 and

150 that have a HCF of 22.

In order to do that we will<u> multiply 22 by the values 5 and 6</u>

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Hence the first two numbers between 100 and 150 that have a HCF of 22 are 110 and 132

Learn more on HCF here; brainly.com/question/21504246

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To answer this problem, one way is to use the calculator and add directly the two numbers and get the average between the two. The answer is 25/77. We can also get this number by getting the LCM of the denominators which is 77. 2/7 is equal to 22/77 while 4/11 is equal to 28/77. 0.5*(22 +28) is equal to 25. Hence the answer is 25/77. 
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