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3 years ago

What is the standard form of the equation of the line through (6,-3) with a slope of 2/3

2 answers:

7 0

Y-y1 = m(x-x1)
y+3 = 2/3(x-6)
y+3 = 2/3 x -4
y = 2/3 x -7
mult both sides by 3 to get rid of the fraction so. 3y=2x - 7

then put in proper order

2x - 3y = 7

tangare [24]3 years ago

5 0

Ax+by=c is the standard form
you are given the slope and a point, so first write the equation in point-slope form:
y-(-3)=2/3(x-6)
expand the equation: y+3=(2/3)x-4
(2/3)x-y=7 is the standard form

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Answer:

See below

Step-by-step explanation:

By using the table 1 attached (See Table 1 attached)

We can perform all the calculations to express both, y as a function of x or x as a function of y.

Let's make first the line relating y as a function of x.

y as a function of x

(y=response variable, x=explanatory variable)

$\bf y=m_{yx}x+b_{yx}$

where

$\bf m_{yx}$ is the slope of the line

$\bf b_{yx}$ is the y-intercept

In this case we use these formulas:

$\bf m_{yx}=\frac{(\sum y)(\sum x)^2-(\sum x)(\sum xy)}{n\sum x^2-(\sum x)^2}$

$\bf b_{yx}=\frac{n\sum xy-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}$

n = 10 is the number of observations taken (pairs x,y)

Note: Be careful not to confuse

$\bf \sum x^2$ with $\bf (\sum x)^2$

Performing our calculations we get:

$\bf m_{yx}=\frac{(83)(95)^2-(95)(923)}{10*1277-(95)^2}=176.6061$

$\bf b_{yx}=\frac{10*923-(95)(83)}{10(1277)-(95)^2}=0.3591$

So the equation of the line that relates y as a function of x is

In order to compute the standard error $\bf S_{yx}$ , we must use Table 2 (See Table 2 attached) and use the definition

$\bf s_{yx}=\sqrt{\frac{(y-y_{est})^2}{n}}$

and we have that standard error when y is a function of x is

$\bf s_{yx}=\sqrt{\frac{39515985}{10}}=1987.8628$

Now, to find the line that relates x as a function of y, we simply switch the roles of x and y in the formulas.

So now we have:

x as a function of y

(x=response variable, y=explanatory variable)

$\bf x=m_{xy}y+b_{xy}$

where

$\bf m_{xy}$ is the slope of the line

$\bf b_{xy}$ is the x-intercept

In this case we use these formulas:

$\bf m_{xy}=\frac{(\sum x)(\sum y)^2-(\sum y)(\sum xy)}{n\sum y^2-(\sum y)^2}$

$\bf b_{xy}=\frac{n\sum xy-(\sum x)(\sum y)}{n(\sum y^2)-(\sum y)^2}$

n = 10 is the number of observations taken (pairs x,y)

Note: Be careful not to confuse

$\bf \sum y^2$ with $\bf (\sum y)^2$

Remark: If you wanted to draw this line in the classical style (the independent variable on the horizontal axis), you would have to swap the axis X and Y)

Computing our values, we get

$\bf m_{xy}=\frac{(95)(83)^2-(83)(923)}{10*743-(83)^2}=1068.1072$

$\bf b_{xy}=\frac{10*923-(95)(83)}{10(743)-(83)^2}=2.4861$

and the line that relates x as a function of y is

To find the standard error $\bf S_{xy}$ we use Table 3 (See Table 3 attached) and the formula

$\bf s_{xy}=\sqrt{\frac{(x-x_{est})^2}{n}}$

and we have that standard error when y is a function of x is

$\bf s_{xy}=\sqrt{\frac{846507757}{10}}=9200.5856$

In both cases the correlation coefficient r is the same and it can be computed with the formula:

$\bf r=\frac{\sum xy}{\sqrt{(\sum x^2)(\sum y^2)}}$

Remark: This formula for r is only true if we assume the correlation is linear. The formula does not hold for other kind of correlations like parabolic, exponential,..., etc.

Computing the correlation coefficient :

$\bf r=\frac{923}{\sqrt{(1277)(743)}}=0.9478$

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4 years ago

What are the answers to all of these questions?

VikaD [51]

We have that

N 11)
a) graph the equation y=x²-3x-10

using a graph tool
see the attached figure

b) Determine the roots of the equations
the roots of the equations are the values for y=0
x²-3x-10=0
x²-3x-10=(x+2)*(x-5)=0
the roots are
x1=-2
x2=5

see the attached figure problem 11

N 12) what is the value of x in the equation?
3x²=7x

3x²=7x-------> 3x²-7x=0--------> x*(3x-7)=0

x1=0
3x-7=0------> 3x=7------> x=7/3

the values of x are
x1=0
x2=7/3---> 2.33
see the attached figure problem 12

N 13) what is the value of x in the equation?
x²-5x=6----------> x²-5x-6=0
x²-5x-6=0-------> (x+1)*(x-6)=0
the values of x are
x1=-1
x2=6

N 14) what is the value of x in the equation?
2x²-5x=12--------> 2x²-5x-12=0

using a graph tool
see the attached figure problem N 14
2x²-5x-12=0----------> (x+3/2)*(x-4)=0

the values of x are
x1=-3/2
x2=4

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3 years ago

Find the slope of the line on the graph.

Rasek [7]

The slope should be -1/2, because the line goes down 1 unit and then right 2 units.

Also it would be y = -1/2x+1 in y=mx+b.

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3 years ago