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wolverine [178]
3 years ago
12

[6.03] What is the value of the x variable in the solution to the following system of equations?

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
8 0
We multiply both equations by a constant term so that when we subtract the equations, the y values cancel out:

4(4x - 3y = 3)
3(5x - 4y = 3)

16x - 12y = 12
15x -12y = 9; subtracting the second equation from the first:

x = 3
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How could you find 3/8% of 800
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A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midni
Alex17521 [72]

Answer: It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

Step-by-step explanation:

Since we have given that

n = 84

x = 11

So, \hat{p}=\dfrac{x}{n}=\dfrac{11}{84}=0.13

p = 0.20

So, hypothesis:

H_0:p=\hat{p}\\\\H_a:\hat{p}

so, test statistic value would be

z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.13-0.20}{\sqrt{\dfrac{0.2\times 0.8}{84}}}\\\\z=-1.604

At 1% level of significance, critical value would be

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Since 2.58>-1.604

So, We will accept the null hypothesis.

Hence, It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

4 0
3 years ago
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Each of 16 students measured the circumference of a tennis ball by four different methods, which were: A: Estimate the circumfer
almond37 [142]

Answer:

Following are the solution to the given equation:

Step-by-step explanation:

Please find the complete question in the attachment file.

In point a:

\to \mu=\frac{\sum xi}{n}

       =22.8

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{119.18}{16-1}}\\\\ =\sqrt{\frac{119.18}{15}}\\\\ = \sqrt{7.94533333}\\\\=2.8187

In point b:

\to \mu=\frac{\sum xi}{n}

       =20.6875  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{26.3375}{16-1}}\\\\=\sqrt{\frac{26.3375}{15}}\\\\ =\sqrt{1.75583333}\\\\ =1.3251

In point c:

 \to \mu=\frac{\sum xi}{n}

         =21  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{2.62}{16-1}}\\\\ =\sqrt{\frac{2.62}{15}} \\\\= \sqrt{0.174666667}\\\\=0.4179

In point d:

\to \mu=\frac{\sum xi}{n}

       =20.8375  

\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}

       =\sqrt{\frac{8.2975}{16-1}}\\\\ =\sqrt{\frac{8.2975}{15}} \\\\  =\sqrt{0.553166667} \\\\ =0.7438

6 0
3 years ago
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