= Solution
The one-day slaughter-inspection yield is
$$
\widehat p_{\mathrm{inspection}}=\frac{15}{30000}=0.0005,
$$
or 500 detections per million inspected adult cattle. The routine field detections correspond, using a stable herd size of five million, to
$$
\widehat r_{\mathrm{field}}=\frac{14600}{5\times10^6\times365}=8.0\times10^{-6}
$$
per adult-cattle day, or 8 detections per million adult-cattle days. Therefore the numerical one-day comparison is
$$
\boxed{\frac{0.0005}{8.0\times10^{-6}}=62.5.}
$$
The annual field rate is $14600/(5\times10^6)=0.00292$ per cattle-year; comparing the inspection <probability> directly with this annual proportion would mix observation periods. Applying the random inspection-sample proportion to all 150000 slaughtered adults would predict about 75 detected cases in that slaughter cohort. These arithmetic comparisons describe detection yields, not a 62.5-fold biological disease-risk difference: a cross-sectional inspection estimates a <prevalence> yield, whereas daily field reporting is an ascertainment rate over animal-time.
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