A. 36 B. 12 C. 80 D. 60 E. 48 F. 20

A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like

mixas84 [53]

Answer:

a) $\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) $\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) the steady state mass of the drug is 2000 mg

d) t ≅ 153.51 minutes

Step-by-step explanation:

From the given information;

At time t= 0

an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of 500

The inflow rate is 0.06 L/min.

Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant.

The objective of the question is to calculate the following :

a) Write an initial value problem that models the mass of the drug in the blood for t ≥ 0.

From above information given :

$Rate _{(in)}= 500 \ mg/L \times 0.06 \ L/min = 30 mg/min$

$Rate _{(out)}=\dfrac{x}{4} \ mg/L \times 0.06 \ L/min = 0.015x \ mg/min$

Therefore;

$\dfrac{dx}{dt} = Rate_{(in)} - Rate_{(out)}$

with respect to x(0) = 0

$\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.

$\dfrac{dx}{dt} = -0.015(x - 2000)$

$\dfrac{dx}{(x - 2000)} = -0.015 \times dt$

By Using Integration Method:

$ln(x - 2000) = -0.015t + C$

$x -2000 = Ce^{(-0.015t)$

$x = 2000 + Ce^{(-0.015t)}$

However; if x(0) = 0 ;

Then

C = -2000

Therefore

$\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) What is the steady-state mass of the drug in the blood?

the steady-state mass of the drug in the blood when t = infinity

$\mathbf{x = 2000 - 2000e^{-0.015 \times \infty }}$

x = 2000 - 0

x = 2000

Thus; the steady state mass of the drug is 2000 mg

d) After how many minutes does the drug mass reach 90% of its stead-state level?

After 90% of its steady state level; the mas of the drug is 90% × 2000

= 0.9 × 2000

= 1800

Hence;

$\mathbf{1800 = 2000 - 2000e^{(-0.015t)}}$

$0.1 = e^{(-0.015t)$

$ln(0.1) = -0.015t$

$t = -\dfrac{In(0.1)}{0.015}$

t = 153.5056729

t ≅ 153.51 minutes

4 0

3 years ago

Sheila had 8 yards of fabric.she used 2/3 of the fabric on a project. How many yards does she have left?

Inessa [10]

Answer:

As a decimal it can be written as 7.3 repeated

5 0

3 years ago

Simplify the following expression 5^2-4/5+2

Sever21 [200]

26 1/5 is the answer i got

8 0

3 years ago

Read 2 more answers

Suppose that the useful life of a car battery, measured in months, decays with parameter 0.03 . We are interested in the life of

scZoUnD [109]

Answer:

a) X ~ $e(0.03)$

b) μ = 100/3

c) $\sigma = 100/3$

d) A battery is expected to last 100/3 months (33 months and 10 days approximately).

e) For seven batteries, i would expect them to last 700/3 months (approximately 19 years, 5 months and 10 days).

Step-by-step explanation:

a) The life of a battery is usually modeled with an exponential distribution X ~ $e(0.03)$

b) The mean of X is μ = 1/0.03 = 100/3

c) The standard deviation is $\sigma = 1/0.03 = 100/3$

d) The expected value of the bateery life is equal to its mean, hence it is 100/3 months.

e) The expected value of 7 (independent) batteries is the sum of the expected values of each one, hence it is 7*100/3 = 700/3 months.

4 0

3 years ago

Wp x 6 = 2 Nees help on this

sdas [7]

Answer:

x = 1/3

Step-by-step explanation:

x * 6 = 2

x = 2/6

x = 1/3

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8 0

3 years ago