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Delvig [45]
3 years ago
14

7x+2y=3 and x-3y=30, what is the value of y?

Mathematics
2 answers:
worty [1.4K]3 years ago
8 0

Answer:

y = -9

x = 3

Step-by-step explanation:

in order to determine the value of y in the equation we would say let

7x+2y=3 .......................................... equation 1

x-3y=30.......................................... equation 2

from equation 2

x-3y=30.......................................... equation

x = 30 + 3y..................................... equation 3

substitute for equation 3 into equation 1

7x+2y=3 .......................................... equation 1

7( 30 + 3y) + 2y = 3

210 + 21y + 2y = 3

210 + 23y = 3

combine the like terms

23y = 3 - 210

23y = - 207

divide both sides by the coefficient of y

23y/23 = -207/23

y = -9

substitute for y =-9 into equation 3

x = 30 + 3y..................................... equation 3

x = 30 + 3(-9)

x = 30 -27

x = 3

Blizzard [7]3 years ago
5 0

\tt\it\bf\it\huge\bm{\mathfrak{{\red{hello*mate♡}}}}

\tt\it\bf\huge\it\bm{\mathcal{\fcolorbox{blue}{yellow}{\red{ANSWER==>}}}}

_______________________________

<h3>I'm solving it using substitution method:-</h3>

<h3>7x+2y=3 {given}</h3>

<h3>=>7x=3-2y</h3>

<h3>=>x=(3-2y)/7-------(1)</h3>

<h3>x-3y=30 {given}</h3>

<h3>=>x=30+3y</h3>

<h3>=>(3-2y)/7=30+3y {putting the value of x from eqn 1}</h3>

<h3>=>3-2y=210+21y</h3>

<h3>=>3-210=21y+2y</h3>

<h3>=>-207=23y</h3>

<h3>=>y= -207/23= -9</h3>

<h3>putting the value of y on eqn(1):-</h3>

<h3>x=(3-2y)/7</h3>

<h3>x=>(3-2(-9))/7=(3+18)/7=21/7=3</h3>

<h2>Hence, x=3, y= -9</h2>

✌️✌️❤️❤️

_______________________________

\tt\it\bf\huge\it\bm{\mathcal{\fcolorbox{blue}{yellow}{\red{BE-BRAINLY !}}}}

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Step-by-step explanation:

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5 0
4 years ago
A researcher is interested in finding a 95% confidence interval for the mean number minutes students are concentrating on their
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Answer:

A. Normal

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Step-by-step explanation:

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

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A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

Question A:

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Question B:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96\frac{12}{\sqrt{150}} = 1.92

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.92 = 40.08 minutes

The upper end of the interval is the sample mean added to M. So it is 42 + 1.92 = 43.92 minutes

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Question C:

x% confidence interval -> x% will contain the true population mean, (100-x)% wont.

So, 95% confidence interval:

About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.

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Hope this helped :)

Have a good one!

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4 0
2 years ago
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