A, the interquartile range is 10, does not fit.
The interquartile range is found by subtracting the upper quartile and the lower quartile; in a box-and-whisker plot these are the outer edges of the box. In this case, they are 40 and 20; 40-20 = 20, not 10.
One line passes through the points \blueD{(-3,-1)}(−3,−1)start color #11accd, (, minus, 3, comma, minus, 1, ), end color #11accd
mart [117]
Answer:
The lines are perpendicular
Step-by-step explanation:
we know that
If two lines are parallel, then their slopes are the same
If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)
Remember that
The formula to calculate the slope between two points is equal to
<em>Find the slope of the first line</em>
we have the points
(-3,-1) and (1,-9)
substitute in the formula
<em>Find the slope of the second line</em>
we have the points
(1,4) and (5,6)
substitute in the formula
Simplify
<em>Compare the slopes</em>
Find out the product

therefore
The lines are perpendicular
Answer:
<h2> </h2><h2>

</h2>
Step-by-step explanation:
<h3>
<u>Question</u><u>:</u><u>-</u></h3>
<h3>
<u>Equation</u><u>:</u><u>-</u></h3>
<h3>
<u>Solution</u><u>:</u><u>-</u></h3>
=> 3x + 4y + 6z = 15
- <em>[</em><em>On</em><em> </em><em>subtracting</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>with</em><em> </em><em>4y</em><em>]</em>
=> 3x + 4y + 6z - 4y = 15 - 4y
- <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>
=> 3x + 6z = 15 - 4y
- <em>[</em><em>On</em><em> </em><em>subtracting</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>with</em><em> </em><em>6z</em><em>]</em>
=> 3x + 6z - 6z = 15 - 4y - 6z
- <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>
=> 3x = 15 - 4y - 6z
- <em>[</em><em>On</em><em> </em><em>dividing</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>with</em><em> </em><em>3</em><em>]</em>

- <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>

C.4 is the answer to the question