Answer:
6
Step-by-step explanation:
Solve the absolutes: |-6| becomes 6
and |6| doesn't change and still is 6
Substitute: 6-6-(-6)
- * - = +
6-6+6 = 6
6/25.2
Multiply both sides by 5 so 25.2 is a whole number:
30/126
and then simplify by dividing by 6:
5/21 which is equal to 0.2381.
Hope this helps :)
the sum of the angles in an ngon (a polygon with n sides) is (n-2)*180
so an octagon has 8 sides and has a sum total of (8-2)*180=1080 degrees
if each of 7 angles is 132, then 1080=?+7*132
1080=?+924
156=?
the 8th angle is 156 degrees
The <em>proposed</em> design of the atrium (<em>V < V'</em>) is possible since its volume is less than the <em>maximum possible</em> atrium.
<h3>Can this atrium be built in the rectangular plot of land?</h3>
The atrium with the <em>maximum allowable</em> radius (<em>R</em>), in feet, is represented in the image attached. The <em>real</em> atrium is possible if and only if the <em>real</em> radius (<em>r</em>) is less than the maximum allowable radius and therefore, the <em>real</em> volume (<em>V</em>), in cubic feet, must be less than than <em>maximum possible</em> volume (<em>V'</em>), in cubic feet.
First, we calculate the volume occupied by the maximum allowable radius:
<em>V' =</em> (8 · π / 3) · (45 ft)³
<em>V' ≈</em> 763407.015 ft³
The <em>proposed</em> design of the atrium (<em>V < V'</em>) is possible since its volume is less than the <em>maximum possible</em> atrium. 
To learn more on volumes, we kindly invite to check this verified question: brainly.com/question/13338592
Assuming that each marble can be picked with equal probability, we notice that there is a total of

marbles, of which 2 are red.
So, the probability of picking a red marble is

In fact, as in any other case of (finite) equidistribution, we used the formula
