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3 years ago

There is a strong positive liner correlation between rainfall....

Mathematics

1 answer:

atroni [7]3 years ago

3 0

I’m pretty sure that a is the answer

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Match each pair of points to the equation of the line that is parallel to the line passing through the points.

Luba_88 [7]

Answer:

$B(5,2) \:\: C(7,-5) \:\: \rightarrow y=-3.5x-15$

$D(11,6)\:\: E(5,9)\:\: \rightarrow y=-0.5x-3$

$H(4,4)\:\: I(8,9) \:\: \rightarrow y=1.25x+4$

$L(5,-7)\:\: M(4,-12)\:\: \rightarrow y=5x+9$

Step-by-step explanation:

We need to match the slope of the function with the slope of the lines connecting the two points given. The slope of the lines are as follows:

$B(5,2) \:\: C(7,-5)=\frac{2--5}{5-7} =-3.5$

$D(11,6)\:\: E(5,9)=\frac{6-9}{11-5} =-0.5$

$H(4,4)\:\: I(8,9)=\frac{4-9}{4-8} =1.25$

$L(5,-7)\:\: M(4,-12)=\frac{-7--12}{5-4} =5$

$F(-7,12)\:\: G(3,-8)=\frac{12--8}{-7-3}=-2$

$J(7,2)\:\:K(-9,8) =\frac{2-8}{7--9} =-0.375$

Now,

the slope of the line BC matches with the slope of y=-3.5x-15.

the slope of the line DE matches with the slope of y=-0.5x-3.

the slope of the line HI matches with the slope of y=1.25x+4.

the slope of the line LM matches with the slope of y=5x+9.

and the slopes of the lines FG and JK do not match with any of the functions given.

Thus,

$B(5,2) \:\: C(7,-5) \:\: \rightarrow y=-3.5x-15$

$D(11,6)\:\: E(5,9)\:\: \rightarrow y=-0.5x-3$

$H(4,4)\:\: I(8,9) \:\: \rightarrow y=1.25x+4$

$L(5,-7)\:\: M(4,-12)\:\: \rightarrow y=5x+9$

4 0

3 years ago

Read 2 more answers

What is the equation of this line in standard form? (-4, -1) & (1/2, 3)

BabaBlast [244]

bearing in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{\frac{1}{2}}~,~\stackrel{y_2}{3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{\frac{1}{2}}-\underset{x_1}{(-4)}}}\implies \cfrac{3+1}{\frac{1}{2}+4}\implies \cfrac{4}{~~\frac{9}{2}~~}\implies \cfrac{4}{1}\cdot \cfrac{2}{9}\implies \cfrac{8}{9}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{8}{9}}[x-\stackrel{x_1}{(-4)}]\implies y+1=\cfrac{8}{9}(x+4) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{9}}{9(y+1)=9\left( \cfrac{8}{9}(x+4) \right)}\implies 9y+9=8(x+4)\implies 9y+9=8x+32 \\\\\\ 9y=8x+23\implies -8x+9y=23\implies 8x-9y=-23$