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Alinara [238K]
3 years ago
10

The half-life of a drug is 4 hours. A patient has been taking the drug on a regular basis for a few months and then discontinues

taking it. The concentration of the drug in a patient's system is 120 ng/mL when the patient stops taking the medication. Complete the table, rounded to the nearest tenth of a ng.

Mathematics
1 answer:
brilliants [131]3 years ago
7 0

Answer:

Step-by-step explanation:

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The distance between the two points is 10.
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The second of two numbers is 6 times the first their sum is 77
Galina-37 [17]
6×11=66
11+66=77
so the first number is 11 and the second is 66. together their sum is 77.
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Identify the digit with the given place value in number 173.514 hundredths
Llana [10]

Answer: We are given the number 173.514

Each of the digits in this number has the following place value:

\begin{gathered} 1\rightarrow\text{ Hundreds} \\ 7\rightarrow\text{ Tens} \\ 3\rightarrow\text{ Ones} \\ \text{.}\rightarrow\text{  Decimal Point} \\ 5\rightarrow\text{ Tenth} \\ 1\rightarrow\text{  Hundreth} \\ 4\rightarrow\text{  Thousandth} \end{gathered}

Secondly, We have to identify the numbers in digit 185.712

\begin{gathered} 1\rightarrow\text{ Thousand} \\ 8\rightarrow\text{ Hundred} \\ 5\rightarrow\text{ Tens} \\ \text{. Decimal point} \\ 7\text{ }\rightarrow\text{Thenth} \\ 1\rightarrow\text{ Hundreth} \\ 2\text{ }\rightarrow\text{ Thousandth} \\  \end{gathered}

5 0
1 year ago
What are the solutions of the quadratic equation p^2+3p-28=40
vivado [14]

Answer:

the answer is  =-3±\sqrt{281}/2

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
The mean preparation fee H&R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use t
astraxan [27]

Answer:

a)0.6192

b)0.7422

c)0.8904

d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

Step-by-step explanation:

Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then

z(p)=\frac{ME*\sqrt{N}}{s } where

  • Me is the margin of error from the mean
  • s is the standard deviation of the population
  • N is the sample size

a.

z(p)=\frac{8*\sqrt{30}}{50 } ≈ 0.8764

by looking z-table corresponding p value is 1-0.3808=0.6192

b.

z(p)=\frac{8*\sqrt{50}}{50 } ≈ 1.1314

by looking z-table corresponding p value is 1-0.2578=0.7422

c.

z(p)=\frac{8*\sqrt{100}}{50 } ≈ 1.6

by looking z-table corresponding p value is 1-0.1096=0.8904

d.

Minimum required sample size for 0.95 probability is

N≥(\frac{z*s}{ME} )^2 where

  • N is the sample size
  • z is the corresponding z-score in 95% probability (1.96)
  • s is the standard deviation (50)
  • ME is the margin of error (8)

then N≥(\frac{1.96*50}{8} )^2 ≈150.6

Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

7 0
3 years ago
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