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inessss [21]
1 year ago
10

Patrick and Brooklyn are making decisions about their bank accounts. Patrick wants to deposit $300 as a principle amount, with a

n interest of 3% compounded quarterly. Brooklyn wants to deposit $300 as the principle amount, with an interest of 5% compounded monthly. Explain which method results in more money after 2 years. Show all work.
Mathematics
2 answers:
irina [24]1 year ago
8 0
For this we will use formula that is letting us to input: interest rate, starting funds, how often intereset rate is implemented, period we are observing. Formula looks like this:

M = S(1+ \frac{i}{cp})^cp*y

where M is money, S is starting funds, "i" is interest rate, cp is compounding period and y is number of years. now we express and calculated for both of them and get

M = 318,479 for Patricks investement.

M = 331,482 for Brooklyn.

Which means Brooklyn's method will pay of more.
Tanya [424]1 year ago
8 0
The equation is F=p(1+i/k)^kt 300(1+0.03/4)^4*2=318.48.. Your turn,,can you?
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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
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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

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\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

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