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katrin2010 [14]
3 years ago
13

There are nine players on a baseball team. How many different batting lineups are possible

Mathematics
2 answers:
jolli1 [7]3 years ago
7 0

Answer:

362,880

Step-by-step explanation:

9*8*7*6*5*4*3*2*1=362,880. You multiply the number of players, but subtract one each time.

ipn [44]3 years ago
6 0

Answer: 362,800

Step-by-step explanation: To determine how many different batting orders are possible for a 9-player softball team, let's start by drawing 9 blanks to represent the 9 spots in the order.

We know that 9 different players can fill the first spot in the order. However, once the first spot is filled, there are only 8 players left to fill the second spot and once the second spot is filled, there are only 7 players left to fill the third spot and so on.

Therefore, based on the counting principal, we have 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 ways for all 9 spots to be filled which simplifies to 362,880.

Therefore, there are 362,880 different batting orders for a 9-player softball team.

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Select all the expressions below which represent a 20% discount off the price of an item that originally costs d dollars.
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B. d - 0.2

20%= 20/100= 0.2

so price= d-0.2

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Anybody know how to do this??
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3 years ago
1. Calculate the value of x in the following diagram. 53 - Y IN Q 5x 18​
777dan777 [17]

Answer:

x = 7

Step-by-step explanation:

QY = 5x

YZ = 18

QZ = 53

Thus:

QY + YZ = QZ (segment addition postulate)

5x + 18 = 53

5x = 53 - 18 (subtraction property of equality)

5x = 35

Divide both sides by 5

x = 7

5 0
2 years ago
If the Math Olympiad Club consists of 11 students, how many different teams of 3 students can be formed for competitions?
vivado [14]

Answer:

165 different teams of 3 students can be formed for competitions

Step-by-step explanation:

Combinations of m elements taken from n in n (m≥n) are called all possible groupings that can be made with the m elements so that:

  • Not all items fit
  • No matter the order
  • Elements are not repeated

That is, a combination is an arrangement of elements where the place or position they occupy within the arrangement does not matter. In a combination it is interesting to form groups and their content.

To calculate the number of combinations, the following expression is applied:

C=\frac{m!}{n!*(m-n)!}

It indicates the combinations of m objects taken from among n objects, where the term "n!" is called "factorial of n" and is the multiplication of all the numbers that go from "n" to 1.

In this case:

  • n: 3
  • m: 11

Replacing:

C=\frac{11!}{3!*(11-3)!}

Solving:

C=\frac{11!}{3!*8!}

being:

  • 3!=3*2*1=6
  • 8!=8*7*6*5*4*3*2*1=40,320
  • 11!=39,916,800

So:

C=\frac{39,916,800}{6*40,320}

C= 165

<u><em>165 different teams of 3 students can be formed for competitions</em></u>

3 0
3 years ago
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