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Ivan
3 years ago
11

Suppose that a​ person's peak blood alcohol level is 0.07 ​(grams per 100​ mL) and that this level decreases by 40​% each hour.

​a) What is the hourly decay​ factor? ​b) Write an exponential function ​B(x)equalsCa Superscript x that models the blood alcohol level after x hours. ​c) Evaluate ​B(2​) and interpret the result.
Mathematics
2 answers:
Fed [463]3 years ago
6 0

Answer:

a) Hourly decay factor = 0.6

b) B(x) = 0.07(0.6)^x

c) B(2) = 0.0252

This result means that the blood alcohol level will be 0.0252 (grams per 100 ml) after 2 hours

Step-by-step explanation:

Since the function is an exponential function, it can be modeled as:

B(x) = C (1 + r)^x

Where the peak blood alcohol level, C = 0.07

Since the blood level decreases by 40%(0.4) every hour, r = -0.4

The hourly decay factor is given by a = 1 + r

a = 1 + (-0.4) = 1 - 0.4

a = 0.6

Therefore, the hourly decay factor = 0.6

b) The exponential function is:

B(x) = Ca^x

Where a = 1 + r = 0.6

C = 0.07

B(x) = 0.07(0.6)^x

c) Evaluate B(2)

Substitute x = 2 into the exponential function gotten in part (b)

B(2) = 0.07(0.6)^2\\B(2) = 0.0252

This result means that the blood alcohol level will be 0.0252 (grams per 100 ml) after 2 hours

Natalija [7]3 years ago
3 0

Answer:

Step-by-step explanation:

Considering a person peak blood is 0.07

It decrease by 4% every hour

a) Using exponential function

f(x) = Ca^x

where,

a = 1 + r

r = -0.40%

Here,

C = 0.07,

a = 1 - 0.04 = 0.06

f(x)=0.07(0.6)^x

<h3>Therefore, hourly decay factor is 0.6</h3>

b)

Here the hourly decay factor is 0.6

B(x)=ca^x\\\\0.07(0.6)^x

c) Evaluate

B(2)=0.07\times (0.6)^2\\\\0.07(0.36)\\\\= 0.0252g

0.0252g or 100mL

<h3>Thus, after 2 hours the blood is 0.0252g or 100mL</h3>
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