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matrenka [14]
3 years ago
14

A patient is given a 100-milligram dosage of a drug that decays exponentially at a rate of 11% per hour. Which equation could be

used to find the time when 15 milligrams of the drug remain?
A. 100 = 15(0.89)^x
B. 100 = 15(0.11)^x
C. 15 = 100(0.89)^x
D. 15 100(0.11)^x
Mathematics
1 answer:
kakasveta [241]3 years ago
6 0
Exponential functions can be expressed as:

f=ir^t, f=final value, i=initial value, r=rate or common ratio, t=term number or "time" when it is fractional...In this case we have:

f=100((100-11)/100)^x

f=100(0.89)^x and we want to solve for when f=15 so

15=100(0.89)^x
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A box contains 24 transistors,4 of which are defective. If 4 are sold at random,find the following probabilities. i. Exactly 2 a
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This is a binomial probability. For i, we will apply the Binomial probability formula

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\begin{gathered} P_x=^nC_x\left(p^x\right?\left(q^{n-x}\right) \\ Where\text{ } \\ P_x=binomial\text{ probability} \\ x=number\text{ of times for a specific outcome with n trials =2} \\ p=\text{ probability of success = }\frac{4}{24}=\frac{1}{6} \\ q=probability\text{ of failure =1-}\frac{1}{6}=\frac{5}{6} \\ ^nC_x=\text{ number of combinations = }^4C_2 \\ n=\text{ number of trials = 4} \end{gathered}

Note that I made the probability of being defective as the probability of success = p

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\begin{gathered} P_x=^4C_2\times\lparen\frac{1}{6})^2\times\lparen\frac{5}{6})^{4-2} \\ P_x=6\times\frac{1}{36}\times\frac{25}{36} \\ P_x=\frac{25}{216} \\ =0.1157 \end{gathered}

Hence the answer is 0.1157

(ii) None is defective becomes

\begin{gathered} \lparen\frac{5}{6})^4=\frac{625}{1296} \\ =0.4823 \end{gathered}

hence the answer is 0.4823

(iii) All are defective

\begin{gathered} \lparen\frac{1}{6})^4=\frac{1}{1296} \\ =0.00077 \end{gathered}

(iv) At least one is defective

This is 1 - probability that none is defective

\begin{gathered} 1-\lparen\frac{5}{6})^4 \\ =1-\frac{625}{1296} \\ =\frac{671}{1296} \\ =0.5177 \end{gathered}

Hence the answer is 0.5177

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