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2 years ago

Solve -4x – 8y = -48 8x + 3y = -34.

Mathematics

1 answer:

Likurg_2 [28]2 years ago

4 0

Answer:

x= -1/2 and -32

y = -10

Step-by-step explanation:

1) 2(-4x-8y)= (-48)2

8x+3y =-34

*to create a way to cancel out the x*

2) -8x-16y= -96

8x+3y= -34

*add them

-8x +8x =0

-16y + 3y = -13y

-96+ -34= -130

3) -13y =-130

isolate y

4) y = -10

*plug in* to find x into both original equations*

5) 8x + 3(-10) = -34. -4x-8(-10)= -48

*follow order of operations to solve for x

6) 8x -30 =-34. -4x +80 = -48

7) 8x =-4. -4x =128

8) x= -1/2. x = -32

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The result for the given normal random variables are as follows;

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<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

$\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}$

Part b.

Consider V(3X + 2Y)

$\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}$

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

$3 X+2 Y \sim N(18,77)$

Thus,

$P(3 X+2 Y < 18)=0.5$

Part d.

Consider P(3X + 2Y < 28)

$Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}$

$\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}$

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

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