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Veseljchak [2.6K]
3 years ago
10

Find the midpoint of A and B where A has coordinates (-3,-5) and B has coordinates (4,4).

Mathematics
1 answer:
snow_lady [41]3 years ago
8 0

Answer:

(\frac{1}{2}, \frac{-1}{2})

Step-by-step explanation:

Use the midpoint formula to solve.

1) Given 2 points: (x_{1}, y_1) and (x_2, y_2)

The midpoint formula would be (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}).

2) Based on this, we can easily find the midpoint of AB.

  • -3 would be x_1\\
  • -5 would be y_1
  • 4 would be x_2  
  • the next 4 would be y_2

3) With this information, we can solve for the midpoint.  

(\frac{-3+4}{2}, \frac{-5+4}{2}) --> (\frac{1}{2}, \frac{-1}{2}) --> This is your answer.

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Derek and Mia place two green marbles and one yellow marble in a bag. Somebody picks a marble out of the bag without looking and
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Answer:

Thus, the expected value of points for Derek and Mia are \dfrac{-1}{9} and \dfrac{1}{9} respectively.

Step-by-step explanation:

Number of green marbles = 2 and Number of Yellow marbles = 1

Then, total number of marbles = 2+1 = 3

A person selects two marbles one after another after replacing them.

So, the probabilities of selecting different combinations of colors are,

1.\ P(GG)=P(G)\times P(G)\\\\P(GG)=\dfrac{2}{3}\times \dfrac{2}{3}\\\\P(GG)=\dfrac{4}{9}

2.\ P(GY)=P(G)\times P(Y)\\\\P(GY)=\dfrac{2}{3}\times \dfrac{1}{3}\\\\P(GY)=\dfrac{2}{9}

3.\ P(YG)=P(Y)\times P(G)\\\\P(YG)=\dfrac{1}{3}\times \dfrac{2}{3}\\\\P(YG)=\dfrac{2}{9}

4.\ P(YY)=P(Y)\times P(Y)\\\\P(YY)=\dfrac{1}{3}\times \dfrac{1}{3}\\\\P(YY)=\dfrac{1}{9}

Now, we have that,

If two marbles are of same color, then Mia gains 1 point and Derek loses 1 point.

If two marbles are of different color, then Derek gains 1 point and Mia loses 1 point.

<h3>Also, the expected value of a random variable X is E(X)=\sum_{i=1}^{n} x_i\times P(x_i).</h3>

Then, the expected value of points for Derek is,

E(D)= (-1)\times \dfrac{4}{9}+1\times \dfrac{2}{9}+1\times \dfrac{2}{9}+(-1)\times \dfrac{1}{9}\\\\E(D)= \dfrac{-5}{9}+\dfrac{4}{9}\\\\E(D)=\dfrac{-1}{9}

And the expected value of points for Mia is,

E(M)= 1\times \dfrac{4}{9}+(-1)\times \dfrac{2}{9}+(-1)\times \dfrac{2}{9}+1\times \dfrac{1}{9}\\\\E(M)= \dfrac{5}{9}-\dfrac{4}{9}\\\\E(M)=\dfrac{1}{9}.

Thus, the expected value of points for Derek and Mia are \dfrac{-1}{9} and \dfrac{1}{9} respectively.

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