Answer:
a. dQ/dt = -kQ
b. ![Q = 9e^{-kt}](https://tex.z-dn.net/?f=Q%20%3D%209e%5E%7B-kt%7D)
c. k = 0.178
d. Q = 1.063 mg
Step-by-step explanation:
a) Write a differential equation for the quantity Q of hydrocodone bitartrate in the body at time t, in hours, since the drug was fully absorbed.
Let Q be the quantity of drug left in the body.
Since the rate of decrease of the quantity of drug -dQ/dt is directly proportional to the quantity of drug left, Q then
-dQ/dt ∝ Q
-dQ/dt = kQ
dQ/dt = -kQ
This is the required differential equation.
b) Solve your differential equation, assuming that at the patient has just absorbed the full 9 mg dose of the drug.
with t = 0, Q(0) = 9 mg
dQ/dt = -kQ
separating the variables, we have
dQ/Q = -kdt
Integrating we have
∫dQ/Q = ∫-kdt
㏑Q = -kt + c
![Q = e^{-kt + c}\\Q = e^{-kt}e^{c}\\Q = Ae^{-kt} (A = e^{c})](https://tex.z-dn.net/?f=Q%20%3D%20e%5E%7B-kt%20%2B%20c%7D%5C%5CQ%20%3D%20e%5E%7B-kt%7De%5E%7Bc%7D%5C%5CQ%20%3D%20Ae%5E%7B-kt%7D%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%28A%20%3D%20e%5E%7Bc%7D%29)
when t = 0, Q = 9
![Q = Ae^{-kt} \\9 = Ae^{-k0}\\9 = Ae^{0}\\9 = A\\A = 9](https://tex.z-dn.net/?f=Q%20%3D%20Ae%5E%7B-kt%7D%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%5C%5C9%20%3D%20Ae%5E%7B-k0%7D%5C%5C9%20%3D%20Ae%5E%7B0%7D%5C%5C9%20%3D%20A%5C%5CA%20%3D%209)
So, ![Q = 9e^{-kt}](https://tex.z-dn.net/?f=Q%20%3D%209e%5E%7B-kt%7D)
c) Use the half-life to find the constant of proportionality k.
At half-life, Q = 9/2 = 4.5 mg and t = 3.9 hours
So,
![Q = 9e^{-kt}\\4.5 = 9e^{-kX3.9}\\\frac{4.5}{9} = e^{-kX3.9}\\\frac{1}{2} = e^{-3.9k}\\](https://tex.z-dn.net/?f=Q%20%3D%209e%5E%7B-kt%7D%5C%5C4.5%20%3D%209e%5E%7B-kX3.9%7D%5C%5C%5Cfrac%7B4.5%7D%7B9%7D%20%3D%20e%5E%7B-kX3.9%7D%5C%5C%5Cfrac%7B1%7D%7B2%7D%20%3D%20e%5E%7B-3.9k%7D%5C%5C)
taking natural logarithm of both sides, we have
d) How much of the 9 mg dose is still in the body after 12 hours?
Since k = 0.178,
![Q = 9e^{-0.178t}](https://tex.z-dn.net/?f=Q%20%3D%209e%5E%7B-0.178t%7D)
when t = 12 hours,
![Q = 9e^{-0.178t}\\Q = 9e^{-0.178X12}\\Q = 9e^{-2.136}\\Q = 9 X 0.1181\\Q = 1.063 mg](https://tex.z-dn.net/?f=Q%20%3D%209e%5E%7B-0.178t%7D%5C%5CQ%20%3D%209e%5E%7B-0.178X12%7D%5C%5CQ%20%3D%209e%5E%7B-2.136%7D%5C%5CQ%20%3D%209%20X%200.1181%5C%5CQ%20%3D%201.063%20mg)
I assume (x-4)3 means (x-4)³. What we wish is to set the derivative equal to zero.
Expanding the T(x) polynomial makes it easier for me to take the derivative.
So (x-4)³ = x³ - 12x² + 48x - 64 + 6
T'(x) = 3x² - 24x + 48
We can factor out a 3 and set this to zero:
x² - 8x + 16 = 0
(x -4)² = 0
x = 4 should therefore represent the turning point.
I am mildly chagrined, I almost used the f'(x) = nx^(n-1) function at first, which appears would have been correct.
Answer:
Step-by-step explanation:
BC=CD
AC=ED
AB=CE
Answer:
To use a paired t-test
Step-by-step explanation:
t-tests are essentially tests carried out to compare two different groups to check if there is a major difference in their performances or results which may be related in certain characteristics.
A paired (correlated) t-test is used when there are matched pair of similar units or instances of repeated measurements.
In this case, Mary has to use a paired t-test because the samples (students) have high similarity in characteristics. The students in question has to be from the same class to test her hypothesis (similar characteristics).