On an American League baseball team with 15 field players and 10 pitchers, the manager must select for the starting line-up, 8 field players, 1 pitcher and 1 designated hitter. A starting line-up specifies the players for these positions and the positions in a batting order for the 8 field players and designated hitter. If the designated hitter must be chosen among all the possible field players, how many possible starting line-ups are there.

There are 18,162,144,000 possible starting line-ups

Step-by-step explanation:

The order in which the player are chosen is important, as if two player exchange places, we have a new batting order. So we use the permutations formula to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

In this question:

Batting order:

9 batters(8 field + the designated hitter) from a set of 15. So

Pitcher:

1 from a set of 10. So

Total:

There are 18,162,144,000 possible starting line-ups

3201

step-by-step explanation:

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation: