What are the steps to find lower and upper quartiles of a data set

1 answer:

5 0

The lower quartile is the value of the middle of the first set.

The upper quartile is the value of the middle of the second set.

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What is the answer for this question 7(x+4)-7=48-2x

Alex777 [14]

Answer:

x = 3

Step-by-step explanation:

7(x + 4) - 7 = 48 - 2x ← distribute parenthesis and simplify left side

7x + 28 - 7 = 48 - 2x

7x + 21 = 48 - 2x ( add 2x to both sides )

9x + 21 = 48 ( subtract 21 from both sides )

9x = 27 ( divide both sides by 9 )

x = 3

8 0

2 years ago

Read 2 more answers

Calculator

Sedaia [141]

Check the picture below.

so let's find the lengths of those two sides in red, since are the length and width of the rectangle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[-3-(-6)]^2+[6-3]^2}\implies d=\sqrt{(-3+6)^2+(6-3)^2} \\\\\\ d=\sqrt{9+9}\implies \boxed{d=\sqrt{18}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-1})~\hfill d=\sqrt{[-2-(-6)]^2+[-1-3]^2} \\\\\\ d=\sqrt{(-2+6)^2+(-1-3)^2}\implies d=\sqrt{16+16}\implies \boxed{d=\sqrt{32}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the rectangle}}{(\sqrt{18})(\sqrt{32})}\implies \sqrt{18\cdot 32}\implies \sqrt{576}\implies 24$