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sammy [17]
4 years ago
12

Someone help me please 30 points

Mathematics
2 answers:
4vir4ik [10]4 years ago
6 0

Answer:

1. -41

2. -40

3. -4

4. -37

5. -24

6. -33

8. -47

9. 14

10. 59

11. -20

12. -16

13. 19


fredd [130]4 years ago
4 0

Answer:4.-37 5.-6 6.-33 11.-20 12.-16

13.19


Step-by-step explanation:


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Answer:
1) 20
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1. (12 points) The tennis team at Taft High School has 8 players, and the tennis team at McKinley High School has 7 players. Reb
Marta_Voda [28]

Complete Question

1. (12 points) The tennis team at Taft High School has 8 players, and the tennis team at McKinley High School has 7 players. Rebecca is on the team at Taft, and her sister Leah is on the team at McKinley. The two teams are going to play a tournament with 4 rounds. In each round, one player from Taft will play a match against one player from McKinley. Each player can play in at most one match. A schedule for the tournament consists of an ordered list of the names of the players from each school who will play in each of the four rounds. The schedule does not include the results of each match.

(a) How many schedules are there?

(b) How many schedules include neither sister?

Answer:

a

   N  =  1,411,200

b

  M  =  302, 400

Step-by-step explanation:

From the question we are told that

   The  number of players in Taft High School is t =  8

    The  number of players in McKinley High School  is k = 7

   The number of rounds is  n  =  4

   The highest number of match each player can play is  w =  1

   

Generally the total number of schedules is  

      N  = ^tP_4 * ^k P_4

      N  = ^8P_4 * ^7 P_4

=>   N  =  \frac{8!}{(8-4)!} * \frac{7!}{(7-4)!}

=>   N  =  \frac{8!}{4!} * \frac{7!}{3!}

=>  N  =  \frac{8*7 *6*5*4*3*2*1}{4*3*2*1} * \frac{7*6*5*4*3*2*1}{3*2*1}

=>  N  =  1,411,200

Generally the  number of schedules that include neither of the sisters is  mathematically represented as

        M= \ ^{t-1}P_4 *\ ^ {k-1} P_4

        M= \ ^{8-1}P_4 *\ ^ {7-1} P_4

=>    M= \ ^{7}P_4 *\ ^ {6} P_4

=>     M  =  \frac{7!}{(7-4)!} * \frac{6!}{(6-4)!}

=>     M  =  \frac{7!}{3!} * \frac{6!}{2!}

=>     M  =  \frac{7 *6*5*4*3*2*1}{3*2*1} * \frac{6*5*4*3*2*1}{2*1}

=>      M  =  302, 400

8 0
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What is the sum of the arithmetic sequence 153, 139, 125, …, if there are 22 terms?
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a_1=153\\d=-14\\n=22

substitute

S_{22}=\dfrac{2\cdot153+(22-1)\cdot(-14)}{2}\cdot22=(306+21\cdot(-14))\cdot11=132

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(Where's the expression?)

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Answer:

Step-by-step explanation:

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