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Consider the task of identifying a 1 cm thick breast cancer that is embedded inside a 4.2 cm thick fibroglandular breast as depicted in Fig.
The cancerous tumor has a cross-sectional area of A=1 mm . Let us assume the beam is monoenergetic with photons of energy 20 keV. The total linear attenuation coefficients for the breast cancer and fibroglandular breast tissue at 20 keV are respectively µcancer=0.844 cm-1 and µfibroglandular=0.802 cm-1. First let's consider the case with no scatter at image receptor. Calculate the local radiographic contrast [i.e. C=|Nt-Nb|/Nb] for this particular imaging task.
Next, suppose that there was a constant S/P = 3 at the image receptor. Calculate the local radiographic constrast as in I but now including the scatter contribution.
Q. Describe the Laws of Sines? Ans. Up to now we have dealt exclusively with right triangles. The Law of Sines and the Law of Cosines are used to solve oblique triangles
Laura paid $17 for a pair of jeans. The ticketed price was 20% off the original price plus the sign on the rack said, "Take an additional 15% off the ticketed price." What was the
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41x + 53y = 135, 53x +41y =147 Ans: 41x + 53 y = 135, 53 x + 41 y = 147 Add the two equations : Solve it, to get ... x + y = 3 -------(1) Subtract : Solve it , to
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