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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

1 year ago

By visual inspection, determine the best-fitting regression model for the data plot below

nikklg [1K]

Answer: D. Quadratic

Step-by-step explanation:

4 0

1 year ago

Read 2 more answers

Paul bakes loaves of bread and bread rolls in the ratio of 2:5. If he bakes 750 bread rolls, how many loaves will he bake? A rec

Zanzabum

Paul bakes 300 bread loaves. I don’t know the question you’re asking for the second question.

6 0

1 year ago

Read 2 more answers

Merrill Lynch Securities and Health Care Retirement Inc. are two large employers in downtown Toledo, Ohio. They are considering

ratelena [41]

Answer:

Step-by-step explanation:

We would determine the mean an standard deviation first

Mean = (107 + 92 + 97 + 95 + 105 + 101 + 91 + 99 + 95 + 104)/10 = 98.6

Standard deviation = √(summation(x - mean)/n

n = 10

Summation(x - mean) = (107 - 98.6)^2 + (92 - 98.6)^2 + (97 - 96.6)^2 + (95 - 98.6)^2 + (105 - 98.6)^2 + (101 - 98.6)^2 + (91 - 98.6)^2 + (99 - 98.6)^2 + (95 - 98.6)^2 + (104 - 98.6)^2 = 274

Standard deviation = √(274/10) = 5.23

The confidence interval is used to determine the range of values that could possibly contain a population parameter (population mean)

Confidence interval is written in the form,

(Point estimate - margin of error, Point estimate + margin of error)

The sample mean, x is the point estimate for the population mean.

We will use the t distribution because the sample size is small and the population standard deviation is not given.

Degree of freedom, df = 10 - 1 = 9

α = 1 - 0.99 = 0.01

z α/2 = 0.01/2 = 0.005

The area to the right of z 0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995. Using the t distribution table,

z = 3.25

Margin of error = z × s/√n

Where

s = sample standard Deviation = 5.23

Margin of error = 3.25 × 5.23/√10 = 5.38

Confidence interval = sample mean(point estimate) ± z × σ/√n

The lower boundary of the confidence interval is

98.6 - 5.38 = 93.22

The upper boundary of the confidence interval is

98.6 + 5.38 = 103.98

We estimate with 99% confidence that the mean weekly child care cost of their employees is between $93.22 and $103.98

Also,

99% of the confidence intervals constructed in this way would contain the true value for the population mean of mean weekly child care cost of their employees.

6 0

2 years ago

A print shop makes bumper stickers for election campaigns. If x stickers are ordered (where x < 10,000), then the price per s

Sergeu [11.5K]

Answer:

P(x) = (0.049x - 0.0000015x²)

Step-by-step explanation:

price per sticker is 0.14 − 0.000002x dollars

total cost of producing the order is 0.091x − 0.0000005x² dollars.

P(x) = profit = Revenue - Cost

Let the number of units of stickers made be x

Revenue = (price per sticker) × (total units sold) = (0.14 − 0.000002x) × (x)

= (0.14x - 0.000002x²) dollars.

Cost of producing x units in the order = (0.091x − 0.0000005x²)

P(x) = 0.14x - 0.000002x² - (0.091x − 0.0000005x²) = 0.14x - 0.091x - 0.000002x² + 0.0000005x²

= (0.049x - 0.0000015x²)

P(x) = (0.049x - 0.0000015x²)

Hope this Helps!!!

3 0

2 years ago

Assemble the proof by dragging tiles to the statements and reasons columns￼

Assemble the proof by dragging tiles to the statements and reasons columns