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Andrej [43]
2 years ago
10

Researchers are recording how much of an experimental medication is in a person's bloodstream every hour. they discover that hal

f-life of the medication is about 6 hours.​

Mathematics
1 answer:
Furkat [3]2 years ago
7 0

By using the known half-life, we can see that if the initial dose is 500mg, after 4 days the medication will be 0.0076 mg.

<h3>What is the half-life?</h3>

We define half-life as the time such that the initial amount is reduced to its half.

So, if a given substance has a half-life T, then the amount of substance as a function of time, we have:

S(t) = A*e^{-t*ln(2)/T}

Where t is the variable in time units.

We know that T = 6 hours, and A, the initial dose, is 500 mg, so the formula is:

S(t) = 500mg*e^{-t*ln(2)/6h}

Then after 4 days (or 4*24h = 96h) the amount of medication is:

S(96h) =  500mg*e^{-96h*ln(2)/6h} = 0.0076 mg.

If the initial dose was 750mg, after 4 days the person would have:

S(96h) = 750mg*e^{-96h*ln(2)/6h} = 0.0114 mg

So the person would have:

0.0114 mg -  0.0076 mg = 0.038 mg more of medication.

If you want to learn more about the half-life, you can read:

brainly.com/question/11152793

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Multiply Both Sides By -1 (Reverse Inequality)

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Read 2 more answers
If A and B are independent events with P(A) = .5 and P(B) = .2, find the following:a) P(A U B)b) P(A^c ? B^c)c) P(A^c U B^c)**No
Effectus [21]

If A and B are independent, then P(A\cap B)=P(A)P(B).

a.

P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)

P(A\cup B)=0.5+0.2-0.5\cdot0.2

\boxed{P(A\cup B)=0.6}

b. I'm guessing the ? is supposed to stand for intersection. We can use DeMorgan's law for complements here:

P(A^c\cap B^c)=P(A\cup B)^c=1-P(A\cup B)

P(A^c\cap B^c)=1-0.6

\boxed{P(A^c\cap B^c)=0.4}

c. DeMorgan's law can be used here too:

P(A^c\cup B^c)=P(A\cap B)^c=1-P(A\cap B)=1-P(A)P(B)

P(A^c\cup B^c)=1-0.5\cdot0.2

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4 0
3 years ago
US average math SAT scores follow a normal distribution with a mean of 505 and a standard deviation of 112. A sample of 64 enter
Hitman42 [59]

Answer:

The claim that the scores of UT students are less than the US average is wrong

Step-by-step explanation:

Given : Sample size = 64

           Standard deviation = 112

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To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.

Solution:

Sample size = 64

n > 30

So we will use z test

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Refer the z table for p value

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α=0.05

p value > α

So, we accept the null hypothesis

Hence The claim that the scores of UT students are less than the US average is wrong

3 0
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