On the following dartboard, the radius of the bulls-eye (area A) is 4 inches. The radius of each concentric circle is 4 inches more than the circle inside it. If a person throws randomly onto the dartboard, what is the probability that the dart will hit in area B? (1 point) Responses Image with alt text: three-sixteenths Image with alt text: one-half Image with alt text: one-fourth
Image with alt text: three-sixteenths
The probability of hitting area B can be calculated by comparing the areas of the circles representing B and the entire dartboard.
The area of the bulls-eye (area A) is π(4)^2 = 16π square inches.
The area of the entire dartboard can be calculated by adding the areas of all the concentric circles:
π(4)^2 + π(8)^2 + π(12)^2 + π(16)^2 + ...
This can be simplified to:
π(4)^2(1 + 4^2 + 8^2 + 12^2 + ...)
= 16π(1 + 16 + 64 + 144 + ...)
Since each concentric circle has a radius 4 inches greater than the previous one, the areas form a geometric sequence with a common ratio of 4. The sum of this infinite series can be calculated with the formula a/(1 - r), where a is the first term and r is the common ratio.
Using this formula, the area of the entire dartboard is:
16π(1/(1 - 4)) = 16π(-1/3) = -16π/3 square inches
Therefore, the probability of hitting area B is:
Area of B/Area of entire dartboard = 16π/3 / -16π/3 = 16π/3 / -16π/3 = 1/3 = three-sixteenths.
Therefore, the correct answer is three-sixteenths.