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nadezda [96]
4 years ago
6

Match the systems of equations to their solutions. Tiles 2x + y = 12 x = 9 − 2y x + 2y = 9 2x + 4y = 20 x + 3y = 16 2x − y = 11

y = 11 − 2x 4x − 3y = -13 y = 10 + x -3x + 3y = 30 2x + y = 11 x − 2y = -7 Pairs x = 2, y = 7 arrowBoth x = 5, y = 2 arrowBoth x = 3, y = 5 arrowBoth x = 7, y = 3 arrowBoth

Mathematics
1 answer:
AlexFokin [52]4 years ago
4 1
We have 
1) <span>systems of equations 1
</span><span>2x + y = 12
x = 9 − 2y
</span>using a graph tool
see the attached figure
the solution is the point ------------> (5,2)

2) systems of equations 2
x + 2y = 9 
<span>2x + 4y = 20
</span>using a graph tool
see the attached figure
the solution is the point -------> has no solution the two lines are parallel

3) systems of equations 3
<span>x + 3y = 16
2x − y = 11
</span>using a graph tool
see the attached figure
the solution is the point ------------> (7,3)

4) systems of equations 4
<span>y = 11 − 2x
4x − 3y = -13
</span>using a graph tool
see the attached figure
the solution is the point ------------> (2,7)

5) systems of equations 5
y = 10 + x -----------> multiplying by 3-----> 3y=30+3x------> -3x + 3y = 30
<span>-3x + 3y = 30
</span>equation 1 and equation 2 are equals--------> the system has no solution 

6) systems of equations 6
<span>2x + y = 11
x − 2y = -7
</span>using a graph tool
see the attached figure
the solution is the point ------------> (3,5)

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Required Information:  

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Answer:

\bar{x} = 1884\\

Step-by-step explanation:

What is Normal Distribution?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.  

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P(X > \bar{x} )= P(Z > \bar{x}) = 0.10\\P(X < \bar{x} )= P(Z < \bar{x}) = 1 - 0.10\\P(X < \bar{x} )= P(Z < \bar{x}) = 0.90\\

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\bar{x} = \mu + z(\sigma) \\\bar{x} = 1500 + 1.28(300)\\\bar{x} = 1500 + 384\\\bar{x} = 1884\\

Therefore, you need to score 1884 in order to  qualify for the scholarship.

How to use z-table?

Step 1:

In the z-table, find the probability value of 0.90 and note down the value of the that row which is 1.2

Step 2:

Then look up at the top of z-table and note down the value of the that column which is 0.08

Step 3:

Finally, note down the intersection of step 1 and step 2 which is 1.28

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