a) P(identified as explosive) equals P(actual explosive & identified as explosive) + P(not explosive & identified as explosive) = (10/(4*10^6))*0.95+(1-10/(4*10^6))*0.005 = 0.005002363. Thus, the probability that it actually contains explosives given that it's identified as containing explosives is (10/(4*10^6))*0.95/0.005002363 = 0.000475. b) Let the probability of correctly identifying a bag without explosives be a. Therefore, a = 0.99999763, approximately 99.999763%. c) No, even if this becomes 1, the true proportion of explosives will always be below half of the total detected.
1.) <span>The ratio of note A compared to middle C is 440.0 / 261.6 = 1.6820 ≈ 1.6818
2.) </span><span>The D note's ratio against middle C is 293.7 / 261.6 = 1.1227
3.) </span><span>D# = 293.6 multiplied by 1.0595 equals 311.1
4.) </span><span>The frequency ratio of G compared to C is 1: 262/392 = 1: 0.6683 ≈ 3: 2
5.) </span><span>The frequency ratio of E to C is 1: 262/330 = 1: 0.7939 ≈ 5: 4
6.) The </span><span>element in a musical note referred to as pitch is the frequency.</span>
