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MakcuM [25]
3 years ago
5

a computer game eliminates half of its players in every round. if the computer game started with 2,300 players, which function m

odels the number of players eliminated in each round
Mathematics
1 answer:
Neko [114]3 years ago
3 0
         We have 2,300 players at the start of the computer game. The game eliminates 1/2 of the players in every round.
        So after the first round we have: 2,300 * 1/2 = 1,150 players left. And after the 2nd round: 2,300 * 1/2 * 1/2 = 2,300 * (1/2)^2 and so on.
       And the function of the players eliminated in each round:
       f ( n ) = 2,300 *  ( (1/2)^1 + (1/2)^2 + ....+ ( 1/2 )^n )
       where n is the number of the round.
       The second part is the sum of the geometric sequence:
       a 1 = 1/2,  q = 1/2:
       S n = 1/2 * (( 1/2)^n - 1 )/ ( 1/2  - 1 ) = - ( ( 1/2)^n - 1 ) =
       = 1 - (1/2)^n
       Answer:  f ( n ) = 2,300 * ( 1 - (1/2)^n ).
      
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Answer:

The value of y given the values of x and z in the question is -734.85

Step-by-step explanation:

This is a joint variation question.

First, we are made to know that y varies jointly as x and z.

The equation for this can be written as;

y = k * x * z

Where k is the constant of proportionality.

Let’s calculate k first.

117 = k * (-3) * (-11)

33k = 117

K = 117/33 = 3.55

Now, we proceed to get what y is when x = -9 and z = 23

Let’s plug these values into the initial equation written.

y = k * x * z

We use the value of k calculated above;

y = 3.55 * (-9) * (23) = -734.85

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