When four basketball players are about to have a free-throw competition, they often draw names out of a hat to randomly select the order in which they shoot. what is the probability that they shoot free throws in alphabetical order? assume each player has a different name.

Step-by-step explanation:

Let the names of 4 players start with letters A, B, C and D.

There is the the only chance that they shoot free throws in alphabetical order (ABCD).

Count how many different ways are to choose 4 letters from the hat. First letter can be chosen in 4 ways, the second letter can be chosen in 3 ways (only 3 letters left), the third letter can be chosen in 2 ways (only 2 letters left) and the fourth letter can be chosen in 1 way (the last letter left). So, there are

different orders.

The probability is

jwjeh; hw; kjhekjhkjhwkejhkw

step-by-step explanation:

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .