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natita [175]
4 years ago
5

Forensic scientists use the following law to determine the time of death of accident or murder victims. If T denotes the tempera

ture of a body t hr after death, then:
T = T0 + (T1 - T0)(0.97)^t
where T0 is the air temperature and T1 is the body temperature at the time of death. John Doe was found murdered at midnight in his house; the room temperature was 67°F and his body temperature was 82°F when he was found. When was he killed? Assume that the normal body temperature is 98.6°F. (Round your answer to two decimal places.)
Mathematics
1 answer:
Naddik [55]4 years ago
7 0

Answer: He was killed 24.83 hours before the body was found.

Step-by-step explanation:

Since we have given that

T=T_0+(T_1-T_0)(0.97)^t

Here, T is the temperature of a body t hr after death, T = 82°F

T₀  is the air temperature , T₀ = 67°F

T₁ is the body temperature at the time of death = T₁ = 98.6°F

So, we will substitute all the values in the above equation.

82=67+(98.6-67)(0.97)^t\\\\82-67=31.6(0.97)^t\\\\\dfrac{15}{31.6}=(0.97)^t\\\\0.474683544=(0.97)^t\\\\\text{Taking ln on both sides}\\\\\ln(0.474683544)=t\ln(0.97)\\\\-0.745=t\times -0.030\\\\t=\dfrac{0.745}{0.030}\\\\t=24.83\ hours

Hence, he was killed 24.83 hours before the body was found.

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