2x – 4y = 14<br> x = 2y – 2<br><br> What is the solution to the system?

The p-value and the alpha value. The p-value refers to the probability which is directly obtainable from the standard score which is referred to as the z-score while the alpha refers to the level of significance.

Now, when the p-value is less than alpha, we simply reject the null hypothesis and accept the alternative hypothesis. In a case however, we have the value of p greater than or equal to the significance level alpha, we simply accept the null hypothesis in this case.

The question asks for a rejection case and this can happen only when the p-value is less than the level of significance alpha

5 0

3 years ago

1.

Eddi Din [679]

Answer:

a) $y=\dfrac{5}{2}x$

b) yes the two lines are perpendicular

c) $y=\dfrac{5}{4}x+6$

Step-by-step explanation:

a) All this is asking if to find a line that is perpendicular to 2x + 5y = 7 AND passes through the origin.

so first we'll find the gradient(or slope) of 2x + 5y = 7, this can be done by simply rearranging this equation to the form $y = mx + c$

$5y = 7 - 2x$

$y = \dfrac{7 - 2x}{5}$

$y = \dfrac{7}{5} - \dfrac{2}{5}x$

$y = -\dfrac{2}{5}x+\dfrac{7}{5}$

this is changed into the $y = mx + c$ , and we easily see that -2/5 is in the place of m, hence $m = \frac{-2}{5}$ is the slope of the line 2x + 5y = 7.

Now, we need to find the slope of its perpendicular. We'll use:

$m_1m_2=-1$ .

here both slopes $m_1$ and $m_2$ are slopes that are perpendicular to each other, so by plugging the value -2/5 we'll find its perpendicular!

$\dfrac{-2}{5}m_2=-1$ .

$m_2=\dfrac{5}{2}$ .

Finally, we can find the equation of the line of the perpendicular using:

$(y-y_1)=m(x-x_1)$

we know that the line passes through origin(0,0) and its slope is 5/2

$(y-0)=\dfrac{5}{2}(x-0)$

$y=\dfrac{5}{2}x$ is the equation of the the line!

b) For this we need to find the slopes of both lines and see whether their product equals -1?

mathematically, we need to see whether $m_1m_2=-1$ ?

the slopes can be easily found through rearranging both equations to $y=mx+c$

Line:1

$2x + 3y =6$

$y =\dfrac{-2x+6}{3}$

$y =\dfrac{-2}{3}x+2$

Line:2

$y = \dfrac{3}{2}x + 4$

this equation is already in the form we need.

the slopes of both equations are

$m_1 = \dfrac{-2}{3}$ and $m_2 = \dfrac{3}{2}$

using

$m_1m_2=-1$

$\dfrac{-2}{3} \times \dfrac{3}{2}=-1$

$-1=-1$

since the product does equal -1, the two lines are indeed perpendicular!

c)if two perpendicular lines have the same intercept, that also means that the two lines meet at that intercept.

we can easily find the slope of the given line, y = − 4 / 5 x + 6 to be $m=\dfrac{-4}{5}$ and the y-intercept is $c=6$ the coordinate at the y-intercept will be (0,6) since this point only lies in the y-axis.