(A) BAC = 0.05
(B) (i) [tex]0.18=\frac{90}{6.8M}[/tex]
(B) (ii) The man's weight is 73.53 kg.
Step-by-step explanation:
Given formula is;
[tex]BAC=\frac{10N-7.5H}{6.8M}[/tex]
(A) A man drinks 4 standard drinks over 2 hours. He weighs 80 kg. Calculate his BAC.
Given,
Standard drinks = N = 4
Time = H = 2
Mass weight = 80 kg
Putting in given formula;
[tex]BAC=\frac{10(4)-7.5(2)}{6.8(80)}\\\\BAC=\frac{40-15}{544}\\\\BAC=\frac{25}{544}\\\\ BAC = 0.046[/tex]
Rounding off to nearest hundredth
BAC = 0.05
(B) A man is picked up with a BAC of 0.18. he drank 12 standard drinks over a period of 4 hours.
(i) substitute the value of N and H in appropriate formula.
Putting values in given formula;
[tex]0.18=\frac{10(12)-7.5(4)}{6.8M}\\\\0.18= \frac{120-30}{6.8M}\\\\0.18=\frac{90}{6.8M}[/tex]
[tex]0.18=\frac{90}{6.8M}[/tex]
(ii) Make M the subject of equation you found in part (b) (i), hence, find the, man's weight.
Equation found in part (i)
0.18 = [tex]\frac{90}{6.8M}[/tex]
Multiplying both sides by M (man's weight) and dividing both sides by 0.18,
[tex]\frac{1}{0.18}*0.18*M=\frac{90}{6.8M}*\frac{1}{0.18}*M\\\\M=\frac{90}{6.8*0.18}\\\\M=\frac{90}{1.224}\\\\M=73.53\ kg[/tex]
The man's weight is 73.53 kg.
Keywords: division, subtraction
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