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xeze [42]
3 years ago
15

Solve the system by the elimination method. Check your work. x - 3y = 0 3y - 6 = 2x

Mathematics
2 answers:
Leto [7]3 years ago
6 0
X - 3y=0
3y - 6 =2x
__________
x - 3y=0 ........(1)
-2x + 3y=6 ........(2)
__________
-x = 6 ==》x = -6 .........(3) subst. in (2)

-2 (-6) + 3y = 6 ==》3y= -6 ==》y = -2

Romashka [77]3 years ago
3 0

Answer:(-6,-2)

Explanation:
The elimination method calls for two variables that cancel each other out. 
In this circumstance, you already have 3y and -3y.

First, make the orders of the equations the same. I got the x's to the same side on each equation.
This left me with:
x=3y        2x=3y-6.
Now, subtract the first equation from the second one to get x=-6.
That's the first part of the solution. For the y, you can just plug x into one of the equations and solve. I'll use the first one. 
-6-3y=0.    
Now get the ys to one side: -6=3y. Divide both sides by 3 to get y= -2.
Now that you have an x and a y coordinate, your solution is (-6,-2).

To check: plug both values in for their variables into each equation.
3(-2)-6=2(-6)   solve to get... -12=-12, which is true, meaning the solution works for that equation.

-6-3(-2)=0   solve to get... 0=0. The solution works for both equations.
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a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean of a population is 74 and the standard deviation is 15.

This means that \mu = 74, \sigma = 15

Question a:

Sample of 36 means that n = 36, s = \frac{15}{\sqrt{36}} = 2.5

This probability is 1 subtracted by the pvalue of Z when X = 78. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

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Question b:

Sample of 150 means that n = 150, s = \frac{15}{\sqrt{150}} = 1.2247

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

Z = \frac{X - \mu}{s}

Z = \frac{77 - 74}{1.2274}

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Z = \frac{X - \mu}{s}

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This probability is the pvalue of Z when X = 74.2. So

Z = \frac{X - \mu}{s}

Z = \frac{74.2 - 74}{1.0136}

Z = 0.2

Z = 0.2 has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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