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Serga [27]
3 years ago
8

Determine the intercepts of the line. − 6 x + 3 y = − 7

Mathematics
1 answer:
egoroff_w [7]3 years ago
7 0

Answer:

Step-by-step explanation:

-6x + 3y = -7

to find the x intercept, sub in 0 for y and solve for x

-6x + 3y = -7

-6x + 3(0) = -7

-6x = -7

x = 7/6 <=== ur x intercept is (7/6,0)

to find the y intercept, sub in 0 for x and solve for y

-6x + 3y = -7

-6(0) + 3y = -7

3y = -7

y = -7/3 <=== ur y intercept is (0,-7/3)

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Use the following table for questions 11-13.A baseball manager believes a linear relationship exists between the number of Home
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The regression equation of Y on X is given by the following formula:

Y-\bar{Y}=b_{yx}(X-\bar{X})

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b_{yx}=\frac{N\sum^{}_{}XY-\sum^{}_{}X\sum^{}_{}Y}{N\sum^{}_{}X^2-(\sum^{}_{}X)^2}

Where N is the number of values (N=8). We need to find the sum of X values, the sum of Y values, the average of X, the average of Y, the sum of X*Y and the sum of X^2.

The table of values is:

The values we need to know are on the following table:

By replacing the known values in the formula we obtain:

\begin{gathered} b_{yx}=\frac{8\cdot26125-167\cdot995}{8\cdot4649-(167)^2} \\ b_{yx}=\frac{209000-166165}{37192-27889} \\ b_{yx}=\frac{42835}{9303} \\ b_{yx}=4.6 \end{gathered}

Now, the average of X and Y is the sum divided by N, then:

\begin{gathered} \bar{X}=\frac{167}{8}=20.87 \\ \bar{Y}=\frac{995}{8}=124.37 \end{gathered}

Replace these values in the formula and find the regression equation as follows:

\begin{gathered} Y-124.37=4.6(X-20.87) \\ Y-124.37=4.6X-4.6\cdot20.87 \\ Y=4.6X-96.11+124.37 \\ Y=4.6X+28.26 \end{gathered}

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Answer:

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Step-by-step explanation:

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3 years ago
Read 2 more answers
Which equation has the solutions x = -3 ± √3i/2 ?
Maurinko [17]

Answer:Answer is option C : [x^{2} + 3x + 3 ] =0

Note:  None of options matches with given question.

instead of "-3" , there should be "-\frac{3}{2}".

Step-by-step explanation:

Note:  None of options matches with given question.

instead of "-3" , there should be "\frac{3}{2}".  

Here, First thing you have to observe the nature of roots.

∴ x = -\frac{3}{2}+\frac{\sqrt{3}}{2}i and x = -\frac{3}{2}-\frac{\sqrt{3}}{2}

∴ [ x+(\frac{3}{2}-\frac{\sqrt{3}}{2}i) ][ x+(\frac{3}{2}+\frac{\sqrt{3}}{2}i) ]=0

∴ [ x^{2} + x(\frac{3}{2}+\frac{\sqrt{3}}{2}i)+ x(\frac{3}{2}-\frac{\sqrt{3}}{2}i) + (\frac{3}{2}-\frac{\sqrt{3}}{2}i)(\frac{3}{2}+\frac{\sqrt{3}}{2}i) ]=0

∴ [x^{2} + \frac{3}{2}x + \frac{\sqrt{3}}{2}ix + \frac{3}{2}x - \frac{\sqrt{3}}{2}ix + (3-\frac{\sqrt{3}}{2}i)(3+\frac{\sqrt{3}}{2}i) ] =0

∴ [x^{2} + 3x + (\frac{3}{2}-\frac{\sqrt{3}}{2}i)(\frac{3}{2}+\frac{\sqrt{3}}{2}i) ] =0

∴ [x^{2} + 3x + \frac{9}{4} - (\frac{\sqrt{3}}{2}i)(\frac{\sqrt{3}}{2}i) ] =0

∴ [x^{2} + 3x + \frac{9}{4} - (\frac{3}{4}) i^{2} ] =0

∴ [x^{2} + 3x + \frac{9}{4} + (\frac{3}{4}) ] =0

∴ [x^{2} + 3x + \frac{12}{4} ] =0  

∴ [x^{2} + 3x + 3 ] =0  

Thus, Answer is option C : <em>[x^{2} + 3x + 3 ] =0  </em>

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