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.
Answer:
600 books
Step-by-step explanation:
The dimensions of the bin are
5 by 2 by 3
The volume of the bin is found by multiplying these three dimensions.
Volume of Bin = 5 * 2 * 3 = 30 cubic feet
To find the volume of each book, we use the same method. The dimensions of one book are:
1 by 0.5 by 0.1
Volume of 1 book = 1 * 0.5 * 0.1 = 0.05 cubic feet
The total number of books fitting into the bin is calculated by:
30/0.05 = 600 books
Let the number be x. The equation x/5 - 17 = 48 leads to (x - 85)/5 = 48. Multiplying gives x - 85 = 240, so x = 240 + 85, resulting in x = 325.
