document.write( "Question 255276: A circular archery target is 122 cm in diameter and consists of a bull's eye of diameter 12.2 cm, surrounded by nine evenly spaced concentric circles. Assume that an arrow that hits a target is equally likely to land anywhere on the target. What's the probability that the arrow will A) hit the bulls eye B) Land somewhere in the outmost four circles. C) Land within the fifth ring in from the outer edge. ---Please help on this problem! THANKS! \n" ); document.write( "

Algebra.Com's Answer #187548 by drk(1908) $\"\"$ $\"About$

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Let's call the bulls eye - circle 1 including circle #1, there are 10 concentric circles.
\n" ); document.write( "radius of circle 1 = 6.1
\n" ); document.write( "radius of circle 2 = 12.2 or 6.1 * 2.
\n" ); document.write( "and so on.
\n" ); document.write( "A) hit the bulls eye
\n" ); document.write( "P(bulls eye) = pi*6.1^2 = 37.21pi cm^2\r
\n" ); document.write( "\n" ); document.write( "B) Land somewhere in the outmost four circles.
\n" ); document.write( "P(outer 4 circles) = p(whole target - inner 6 circles) = pi*R^2 - pi*r^2 = pi*61^2 -pi*36.6^2.
\n" ); document.write( "The 36.6 is the radius of the first 6 circles or 6.1 * 6.
\n" ); document.write( "This gives us 2381.44pi cm^2\r
\n" ); document.write( "\n" ); document.write( "C) Land within the fifth ring in from the outer edge.
\n" ); document.write( "P(5th ring) = p(5th ring - p(4th ring). We want only the fifth ring so all other inner rings must be subtracted out. We get pi*R^2 - pi*r^2 = pi*30.5^2 -pi*24.4^2 = 334.89pi cm^2 \n" ); document.write( "

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