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Brilliant_brown [7]
2 years ago
15

-3x - 2y =-1 4x + 6y =-12 Solve the system of equations

Mathematics
2 answers:
Crank2 years ago
8 0

Answer:

Answer

0

lchavez31

   Expert

   148 answers

   4.5K people helped

Answer:

x=0

Step-by-step explanation:

y=-5x -6

10x + 2y = -12

Substitute y into the equation:

   10x+2(5x-6)=-12

Use distributive property

   10x+10x-12=-12

Add like terms

   20x-12=-12

Add 12 to  both sides of the equation

   20x=0

divide each side by 0

   x=0

Diano4ka-milaya [45]2 years ago
3 0

Answer:

x=3,y=-4

Step-by-step explanation:

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Answer:

a) P(5

And we can find this probability with this difference:

P(-1.90

And using the norma standard distribution or excel we got:

P(-1.90

b) P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)

And using the complement rule we got:

P(Z>-1.096) =1-P(Z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(7.37,1.25)  

Where \mu=7.37 and \sigma=1.25

We are interested on this probability

P(5

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(5

And we can find this probability with this difference:

P(-1.90

And using the norma standard distribution or excel we got:

P(-1.90

Part b

For this case we want this probability:

P(X>6)

And we can use the z score and we got:

P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)

And using the complement rule we got:

P(Z>-1.096) =1-P(Z

8 0
3 years ago
How to evaluate the limit
anzhelika [568]
\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}

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Perhaps you meant to write that x\to-2 instead? In that case, you would have

\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5
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