Problema Solution
The barometric air pressure P in inches of mercury at a distance of d miles from the eye of a severe hurricane can sometes be modeled by the formula P(d)=0.48In(d+1)+27 average air pressure is about 30 inches of mercury.
Evaluate P(0) and P(50). Interpret the results
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P(d)=0.48In(d+1)+27
Find P(0) and P(50) and interpret
P(0)= barometric pressure at 0 miles from the eye of the hurricane
For P(0) plug in 0 for d and solve.
P(0)=0.48ln(0+1)+27
P(0)=0.48ln(1)+27
P(0)=0.48*0+27 ln(1)=0
P(0)=27
P(50)= barometric pressure at 50 miles from eye of the hurricane
For P(50) plug in 50 for d and solve.
P(50)=0.48ln(50+1)+27
P(50)=0.48ln(51)+27
P(50)=0.48*3.9318+27 ln(51)=3.9318
P(50)= 28.89
Interpretation: This means that at 0 miles from the eye of the hurricane, the barometric pressure is 27 inches of mercury, and at 50 miles from the eye, the barometric pressure is 28.89 inches of mercury. Therefore, the barometric pressure is slightly greater further from the hurricanes eye.
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