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FinnZ [79.3K]
3 years ago
5

Solve the system of equations.

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
5 0
(1) 2x-y+z=-7
(2) x-3y+4z=-19
(3) -x+4y-3z=18

Using the method of elimination:
Adding equation (2) and (3):
(x-3y+4z)+(-x+4y-3z)=(-19)+(18)
x-3y+4z-x+4y-3z=-19+18
y+z=-1
Solving for z:
y+z-y=-1-y
z=-1-y

Multiplying the third equation by 2:
(3) -x+4y-3z=18
2(-x+4y-3z=18)
-2x+8y-6z=36
Adding with equation (1)
(2x-y+z)+(-2x+8y-6z)=(-7)+(36)
2x-y+z-2x+8y-6z=-7+36
7y-5z=29
Replacing z=-1-y in the equation above:
7y-5(-1-y)=29
7y+5+5y=29
12y+5=29

Solving for y. Subtracting 5 both sides of the equation.
12y+5-5=29-5
12y=24
Dividing both side of the equation by 12:
12y/12=24/12
y=2

Replacing y=2 in z=-1-y
z=-1-2→z=-3

Replacing y=2 and z=-3 in the equation (2); and solving for x:
(2) x-3y+4z=-19
x-3(2)+4(-3)=-19
x-6-12=-19
x-18=-19
Adding 18 both sides of the equation:
x-18+18=-19+18
x=-1

There is one solution: (x,y,z)=(-1,2,-3)

Answer: Option <span>D. There is one solution (–1, 2, –3).</span>
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An SRS of 25 recent birth records at the local hospital was selected. In the sample, the average birth weight was x = 119.6 ounc
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Answer:

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\mu_{\bar X}= \mu = 119.6

And now for the deviation we have this:

SE_{\bar X} = \frac{6.5}{\sqrt{25}}=1.3

So then the correct answer for this caee would be:

c. 1.30 ounces.

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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\mu_{\bar X}= \mu = 119.6

And now for the deviation we have this:

SE_{\bar X} = \frac{6.5}{\sqrt{25}}=1.3

So then the correct answer for this caee would be:

c. 1.30 ounces.

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