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Leokris [45]
2 years ago
7

Eliminate the x and y

Mathematics
1 answer:
sp2606 [1]2 years ago
7 0

Step-by-step explanation:

Eliminate the x:

\left\{\begin{array}{ccc}x-2y=6\\3x+y=4\end{array}\right\\\\\text{Multiply by (-3)}\\\\\left\{\begin{array}{ccc}-3(x-2y)=(-3)(6)\\3x+y=4\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}-3x+6y=-18\\3x+y=4\end{array}\right}\\.\qquad\qquad7y=-14\\.\qquad\qquad y=-2

Eliminate the y:

\left\{\begin{array}{ccc}x-2y=6\\3x+y=4\end{array}\right\\\\\text{Multiply by 2}\\\\\left\{\begin{array}{ccc}x-2y=6\\2(3x+y)=(2)(4)\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}x-2y=6\\6x+2y=8\end{array}\right}\\\\.\qquad7x=14\\.\qquad x=2

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The means and mean absolute deviations of Sidney’s and Phil’s grades are shown in the table below which expression represents th
Shalnov [3]

The ratio of the difference of the two means to Sidney’s mean absolute deviation is; 4/3.28

<h3>How to find the Mean Absolute Deviation?</h3>

From the given table, we see that;

Mean grade of Sidney = 82

Mean grade of Phil = 78.

Mean absolute deviation of Sidney = 3.28

Mean absolute deviation of Phil = 3.96.

The difference between the two means of Sidney and Phil = 82 - 78 = 4.

Thus, the ratio of the difference of the two means to Sidney’s mean absolute deviation is; 4/3.28

Complete Question is;

The means and mean absolute deviations of Sidney’s and Phil’s grades are shown in the table below. Means and Mean Absolute Deviations of Sidney’s and Phil’s Grades Sidney Phil Mean 82 78 Mean Absolute Deviation 3.28 3.96 Which expression represents the ratio of the difference of the two means to Sidney’s mean absolute deviation?

Read more about Mean Absolute Deviation at; brainly.com/question/447169

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6 0
2 years ago
Can anybody help me out with this?
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5 0
3 years ago
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15. Find the lengths of major and minor arcs AB correct to the tenth of a cm. The minor
oee [108]

Las funciones trigonométricas se utilizan fundamentalmente en la solución de triángulos

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5 0
3 years ago
The NCAA is interested in estimating the difference in mean number of daily training hours for men and women athletes on college
hammer [34]

Answer:

s^2_p = \frac{9*0.3^2 +9*0.4^2}{10+10-2}=0.125

s_p =\sqrt{0.125}=0.354

(2.7 -2.4) - 2.1*0.354\sqrt{\frac{1}{10}+\frac{1}{10}}=-0.032  

(2.7 -2.4)+ 2.1*0.354\sqrt{\frac{1}{10}+\frac{1}{10}}=0.632  

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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".

Data given

\bar X_M = 2.7 represent the sample mean for men

\bar X_F = 2.4 represent the sample mean for women

s_M = 0.3 represent the sample deviation for men

s_F = 0.4 represent the sample deviation for women

n_M = 10 sample size of male

n_F =10 sample size of women

The confidence interval is given by:

(\bar X_M -\bar X_F) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_M}+\frac{1}{n_F}}   (1)

The polled variance can be calculated with this formula:

s^2_p = \frac{9*0.3^2 +9*0.4^2}{10+10-2}=0.125

s_p =\sqrt{0.125}=0.354

For a confidence of 95% the value for the significance is \alpha=1-0.95=0.05 and \alpha/2 = 0.025, the degrees of freedom are given by:

df = n_M + n_F -2= 10+10-2=18

And the critical value can be calculated with the following formula in excel: "=T.INV(1-0.025,18)" and we got t_{\alpha/2}=2.1

Now we can replace into the confidence interval:

(2.7 -2.4) - 2.1*0.354\sqrt{\frac{1}{10}+\frac{1}{10}}=-0.032  

(2.7 -2.4)+ 2.1*0.354\sqrt{\frac{1}{10}+\frac{1}{10}}=0.632  

6 0
3 years ago
Which equations have infinitely many solutions?
igomit [66]

Answer:

4(x - 1) = 4x - 4

3x + 6 = 3(x + 2)

Step-by-step explanation:

The first equation is

5x - 1 = 5x + 1

We simplify to get;

- 1 = 1

This is not true, therefore this equation has no solution.

The second equation is

3x - 1 = 1 - 3x

Combine like terms:

3x + 3x = 1 + 1

6x = 2

x =  \frac{1}{3}

This has a unique solution.

The 3rd equation is

7x - 2 = 2x - 7

Group similar terms:

7x - 2x =  7 + 2 \\ 5x = 9 \\ x  =  \frac{9}{5}

The 4th equation is :

4(x - 1) = 4x - 4

4x - 4= 4x - 4 \\ x = x

This is always true. The equation has infinite solution.

The 5th equation is:

3x + 6 = 3(x + 2) \\ 3x + 6 = 3x + 6 \\ x = x

This also has infinite solution

The 6th equation is

3(x - 4) = 4(x - 3) \\ 3x - 12 = 4x - 12 \\ 3x - 4x =  - 12 + 12 \\ x = 0

It has a unique solution.

8 0
3 years ago
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