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2 years ago

HELP ME PLEASE! I need to find the Interquartile range of : 0,0,0,1,1,3,3,3,4,4,4,6,6,7,7,7,8,8,9,10

Mathematics

1 answer:

Lana71 [14]2 years ago

4 0

Answer:

The interquartile range is 5

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Write an equation in standard form of the line passing through the points (3,3) and (-3,5)

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Answer:

The equation of line AB with points (3,3) and (-3,5) is given as

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Step-by-step explanation:

Here, the given points are A (3, 3) and B (-3,5).

Now, slope of any line is given as :

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or, $m = \frac{5-3}{-3 - 3} = \frac{2}{-6} = -\frac{1}{3}$

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Now , A POINT SLOPE FORM of an equation is

(y - y0) = m (x - x0) ; (x0, y0) is any arbitrary point on line.

So, for the point (3,3) the equation of the line is

y - 3 $y-3 = -\frac{1}{3} (x-3) \implies 3y - 9 = 3 -x$

Hence, the equation of line AB with points (3,3) and (-3,5) is given as:

x + 3y = 12

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3 years ago

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Not sure if that’s right but I hope that helps

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Check the picture below.

$\bf \textit{Law of Cosines}\\ \quad \\
c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies 
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\\\\\\
\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}=cos(C)\implies cos^{-1}\left(\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}\right)=\measuredangle C\\\\
-------------------------------\\\\
\begin{cases}
d=10\\
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