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2 years ago

Find an equation of a line through (2, 1) and perpendicular to - 2x + 4y = 8.

1 answer:

6 0

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$-2x+4y=8\implies 4y=2x+8\implies y=\cfrac{2x+8}{4} \\\\\\ y=\cfrac{2x}{4}+\cfrac{8}{4}\implies y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{2}}x+2\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

since we know that's its slope, then

$\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{1}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{1}\implies -2}}$

so then we're really looking for the equation of a line whose slope is -2 and passes through (2 , 1)

$(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad \qquad \stackrel{slope}{m}\implies -2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{-2}(x-\stackrel{x_1}{2}) \\\\\\ y-1=-2x-4\implies y=-2x-3$

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Consider a value to be significantly low if its z score less than or equal to minus−2 or consider a value to be significantly hi

ziro4ka [17]

Answer:

If the weight is higher than 5.8886 gr would be considered significantly high

If the weight is lower than 5.6121 gr would be considered significantly low

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(5.75241,0.06281)$

Where $\mu=5.75241$ and $\sigma=0.06281$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

For the case when z =-2 we can do this:

$-2 = \frac{X-5.75241}{0.06281}$

And if we solve for X we got:

$X = 5.75241 -2*0.06281 =5.6121$

And for the other case when Z=2 we have:

$2 = \frac{X-5.75241}{0.06281}$

And if we solve for X we got:

$X = 5.75241 +2*0.06281 =5.8886$

If the weight is higher than 5.8886 gr would be considered significantly high

If the weight is lower than 5.6121 gr would be considered significantly low

5 0

3 years ago

Madilyn and Penelope have 28 toys to build. Madilyn builds 3/4 of the toys, while Penelope builds 1/7. How many toys are there l

uysha [10]

Answer:

Step-by-step explanation:

3/4 x 28=21 Madilyn built 21 toys.

1/7 x 28=4 Penelope built 4 toys.

21+4=25 25 toys have been built altogether.

28-25=3

There are 3 toys left to build.

3 0

3 years ago

Please help me I can't fail! Okay, So I have this big algebra test that's worth 30% of my grade. Right now I have a 62 I need a

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Where? It won’t show!

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3 years ago

Gabrielle wants to tile a room with an area of 320 square feet. The width of the room is 4/5 it's length. What are the length an

vampirchik [111]

I hope this helps you

8 0

3 years ago

Elise walks diagonally from one corner of a square plaza to another. Each side of the plaza is 50 meters. What is the diagonal d

jolli1 [7]

Answer:

70.7 meters.

Step-by-step explanation:

We have been given that Elise walks diagonally from one corner of a square plaza to another. Each side of the plaza is 50 meters.

Since we know that diagonal of a square is product of side length of square and $\sqrt{2}$ . So we will find diagonal of our given square plaza by multiplying 50 by $\sqrt{2}$ .

$\text{Diagonal distance across the plaza}=50\times \sqrt{2}$

$\text{Diagonal distance across the plaza}=50\times 1.414213562373095$

$\text{Diagonal distance across the plaza}=70.71067811865475\approx 70.7$

Therefore, diagonal distance across the plaza is 70.7 meters.

8 0

3 years ago