Answer:
2
Step-by-step explanation:
the mode is the one that appears the most, most common one
Answer:

Step-by-step explanation:
The <u>width</u> of a square is its <u>side length</u>.
The <u>width</u> of a circle is its <u>diameter</u>.
Therefore, the largest possible circle that can be cut out from a square is a circle whose <u>diameter</u> is <u>equal in length</u> to the <u>side length</u> of the square.
<u>Formulas</u>



If the diameter is equal to the side length of the square, then:

Therefore:

So the ratio of the area of the circle to the original square is:

Given:
- side length (s) = 6 in
- radius (r) = 6 ÷ 2 = 3 in


Ratio of circle to square:

A = p + prt
The above equation can be solved for “r” in such a
way.
Subtract “p” from both sides;
A – p = p + prt – p
A – p = prt
Divide both sides by “p” and “t”;
A/pt – p/pt = prt/pt
A/pt – 1/t = r
Or
<span>r = A/pt – 1/t</span>
Step-by-step explanation:
4x hope this explanation will help