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A 27-year-old, 5 ft 8 in, 130 lb female is pulled over for driving erratically and fails her field sobriety tests. She is given a roadside evidentiary breath test and has a BAC of 0.15 g/dL. She claims she was drinking with friends earlier in the day but stopped two hours ago. She thought she had consumed two forty-ounce containers of malt liquor (6.2% ethanol by volume) before she stopped drinking.

1. Is her BAC consistent with her story?
2. How many drinks were in her system at the time of the breath test?
3. What should her BAC be based on her statement?

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Answer :

Final answer:

The woman's BAC is consistent with her story. Based on her statement, her BAC should be approximately 0.039 g/dL.

Explanation:

To determine if the woman's BAC is consistent with her story and calculate the number of drinks in her system at the time of the breath test, we can use the Widmark formula. The formula is as follows:

BAC = (Alcohol consumed in grams / (Body weight in grams x Widmark factor)) x 100

First, we need to convert the woman's weight from pounds to grams. Since 1 pound is approximately 453.592 grams, her weight in grams is calculated as:

Weight in grams = 130 lb x 453.592 g/lb

= 58966.96 g

Next, we need to calculate the number of grams of alcohol consumed. The woman claims to have consumed two forty ounce containers of malt liquor. Since 1 ounce is approximately 29.5735 milliliters (ml), the total volume of alcohol consumed is:

Total volume of alcohol consumed = 2 x 40 oz x 29.5735 ml/oz

= 2365.88 ml

Since the malt liquor has an ethanol content of 6.2% by volume, the amount of alcohol in grams is:

Alcohol consumed in grams = 2365.88 ml x 6.2 g/100 ml

= 146.48 g

Now, we can calculate the woman's BAC based on her statement:

BAC = (146.48 g / (58966.96 g x Widmark factor)) x 100

However, we need to determine the Widmark factor, which takes into account the time elapsed since drinking. The average Widmark factor is 0.68 for women. Since the woman claims to have stopped drinking two hours ago, we can use this value:

BAC = (146.48 g / (58966.96 g x 0.68)) x 100

= 0.039 g/dL

Therefore, her BAC based on her statement should be approximately 0.039 g/dL.

Learn more about blood alcohol concentration (bac) calculation here:

https://brainly.com/question/10859601

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