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mr Goodwill [35]
3 years ago
15

Suppose you are running a racing tournament with 12 horses. All eligible horses run in the race at once, and after the first rac

e, the last two horses are eliminated.
a) Find the total number of possible finishes of the first race.

b) How many different possible groups are there for the second race?
(Show work)
Mathematics
1 answer:
borishaifa [10]3 years ago
4 0

Answer:

a)  66

b)  45

Step-by-step explanation:

We have 12 horses and after the first race 2 of them have to be eliminated

that condition is equal to consider that 10 horses will be available for the second race.

The number of possiblities we have for the first race is the combination of 12 horses in group of 10

C¹²₁₀   =  12!/ (2!)* (12-2)!   =  12*11/2   = 66

For the second race

C¹⁰₈  = 10! / 8! * (10-8)!   =   10*9/ 2   = 45

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What is the solution to 3-3x + 9| = -18?
solmaris [256]

Answer:

x = 5 or x = 1

Step-by-step explanation:

Absolute Value Equation entered :

     3|3x-9|=18

Step by step solution :

Step 1: Rearrange this Absolute Value Equation

Absolute value equalitiy entered

     3|3x-9| = 18

Step 2: Clear the Absolute Value Bars

Clear the absolute-value bars by splitting the equation into its two cases, one for the Positive case and the other for the Negative case.

The Absolute Value term is 3|3x-9|

For the Negative case we'll use -3(3x-9)

For the Positive case we'll use 3(3x-9)

Step 3: Solve the Negative Case

     -3(3x-9) = 18

Multiply

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    Rearrange and Add up

     -9x = -9

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     -x = -1

    Multiply both sides by (-1)

     x = 1

    Which is the solution for the Negative Case

Step 4: Solve the Positive Case

(3x-9) = 18

    Multiply

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    Rearrange and Add up

     9x = 45

    Divide both sides by 9

     x = 5

    Which is the solution for the Positive Case

Step 5:  Wrap up the solution

x=1

x=5

Solutions on the Number Line

 

Two solutions were found :

x=5

x=1

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