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fomenos
3 years ago
10

Robin and Evelyn are playing a target game. The object of the game is to get an object as close to the center as possible. Each

player’s score is the number of centimeters away from the center. Robin’s mean is 107, and Evelyn’s mean is 138. Compare the means. Explain what this comparison indicates in the context of the data. Who is winning the game? Why?
ANSWER BELOW for 1st one:
Eveyn's Robins
means : 138 means: 107

So Eveyns mean is greater than Robins beacuse 138 is greater than 107

- THXS FOR LOOKING AND ANOTHER ANSWER IF YOU DONT WANT TO USE THE FIRST ONE PLUSE THEY ARE BOTH RIGHT
Answer for 2 one : Evelyn has the greater mean. It indicates that, on average, her objects landed farther away from the center than Robin’s. This difference means that Robin is winning the game.
Mathematics
1 answer:
denis23 [38]3 years ago
7 0

Answer:

Robin is winning the game and the explanation for the answer is :

Step-by-step explanation:

core represents the number of centimeter each player is away from the center. Given the mean of Robin = 107 and mean of Evelyn = 138

And, as it is mentioned that the player whose object lands more close to the center wins the game.

Evelyn has the greater mean. It indicates that, on average, Evelyn objects landed farther away from the center than Robin's.

Therefore, this difference means that Robin is winning the game.

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2 years ago
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Answer:

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Step-by-step explanation:

Assume your diagram is like the one below.

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For your diagram, that means

\begin{array}{rcl}m\angle L &=&\dfrac{1}{2} \left(m \widehat {JM} - m\widehat {PQ}\right)\\\\(3x + 13)^{\circ}& = &\dfrac{1}{2} \left[(8x + 48)^{\circ} - (5x - 20)^{\circ}\right]\\\\3x + 13& = &\dfrac{1}{2}(8x + 48 - 5x + 20)\\\\3x + 13& = &\dfrac{1}{2}(3x + 68)\\\\6x + 26 & = & 3x + 68\\6x & = & 3x + 42\\3x & = & 42\\x & = & \mathbf{14}\\\end{array}

Check:

\begin{array}{rcl}(3\times14 + 13) & = &\dfrac{1}{2} \left[(8\times14 + 48)^{\circ} - (5\times14 - 20)^{\circ}\right]\\\\42 + 13& = &\dfrac{1}{2}(112 + 48 - 70 + 20)\\\\55& = &\dfrac{1}{2}(110)\\\\55 & = & 55\\\end{array}

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