Problema Solution
5. Essay; show all work. A simple dartboard has three areas… the main board has a radius of 11 inches, there is a circle with a radius of 5 inches, and the bullseye has a radius of 3 inches. What is the probability of a random dart landing inside the bullseye?
Round to the nearest thousandth.
Answer provided by our tutors
P(hitting bullseye) = Area of bullseye region / Area of dartboard
(This assumes a random dart will land somewhere on the dartboard)
= π(3^2) / π(11^2)
= 9/121
= 0.0743
logic and concept=
the Area of Bulls-eye
A = pi *r^2 Start with the area of a circle formula.
A = pi*(3)^2 we plug in r=3
A = 3.14159*(3)^2 we replace pi with 3.14159 ( since pi is approximately 3.14159
A = 3.14159*(9) we square 3 to get 9
A=12.56636 we multiply 3.14159 and 4 to get 12.56636
So the area of the bulls-eye with a radius of 3 units is roughly 28.26square units, the nearest thousandth.
Area of a whole board
A = pi * r^2 Start with the area of a circle formula.
A = pi * (16)^2 we plug in r=16
A = 3.14159*(16)^2 We replace pi with 3.14159 (since pi is approximately 3.14159)
A = 3.14159*(225) Square 16 to get 144.
A=452.16 Multiply 3.14159 and 144 to get
So the area of the circle with a radius of 15units is roughly 452.16 square units.
CHANCE FOR A BULL'S EYE. 28.26/452.16=0.0743 square units