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Sliva [168]
3 years ago
8

A cup of coffee contains about 100 mg of caffeine. Caffeine is metabolized and leaves the body at a continuous rate of about 12%

every hour. a. Write a differential equation for the amount, A, of the caffeine in the body as a function of the number of hours, t, since the coffee was consumed. b. Use the differential equation to find dA/dt at the start of the first hour (right after the coffee is consumed). Use your answer to estimate the change in the amount of caffeine during the first hour.
Mathematics
1 answer:
Mamont248 [21]3 years ago
3 0

Answer:

a. A = C_{0}(1-x)^t\\x: percentage\ of \ caffeine\ metabolized\\

b. \frac{dA}{dt}= -11.25 \frac{mg}{h}

Step-by-step explanation:

First, we need tot find a general expression for the amount of caffeine remaining in the body after certain time. As the problem states that every hour x percent of caffeine leaves the body, we must substract that percentage from the initial quantity of caffeine, by each hour passing. That expression would be:

A= C_{0}(1-x)^t\\t: time \ in \ hours\\x: percentage \ of \ caffeine\ metabolized\\

Then, to find the amount of caffeine metabolized per hour, we need to differentiate the previous equation. Following the differentiation rules we get:

\frac{dA}{dt} =C_{0}(1-x)^t \ln (1-x)\\\frac{dA}{dt} =100*0.88\ln(0.88)\\\frac{dA}{dt} =-11.25 \frac{mg}{h}

The rate is negative as it represents the amount of caffeine leaving the body at certain time.

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Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

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Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

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(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

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