Answer:
C. It is likely
Step-by-step explanation:
Given
See attachment for diagram
For the square.
Let
[tex]Length = L[/tex]
So, the area is:
[tex]Area = Length^2[/tex]
[tex]A_1=L^2[/tex]
The side length of the square equals the diameter of the circle.
So:
[tex]diameter (d) = L[/tex]
The radius is:
[tex]r = \frac{1}{2}d[/tex]
[tex]r = \frac{1}{2}L[/tex]
The area is:
[tex]Area = \pi r^2[/tex]
[tex]A_2 = \pi * (\frac{1}{2}L)^2[/tex]
[tex]A_2 = \pi * \frac{1}{4}L^2[/tex]
[tex]A_2 = \frac{22}{7} * \frac{1}{4}L^2[/tex]
[tex]A_2 = \frac{22}{28} L^2[/tex]
The likelihood that a point will be common to the square and the circle is:
[tex]Pr = \frac{A_2}{A_1}[/tex]
[tex]Pr = \frac{\frac{22}{28} L^2}{L^2}[/tex]
[tex]Pr = \frac{22}{28}[/tex]
[tex]Pr = 78.57\%[/tex]
The above probability is greater than 50% but less than 100%.
Hence, it is likely
Answer:
C. Likely
Step-by-step explanation:
all the points are in the circle so it's likely to choose a point inside the circle
