Question

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.

Answer 1

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

$f(x)=0.07(0.6)^x$

Therefore, hourly decay factor is 0.6

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

Thus, after 2 hours the blood is 0.0252g or 100mL

Answer 2

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