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oksian1 [2.3K]
4 years ago
14

Grace starts with 100 milligrams of a radioactive substance. The amount of the

Mathematics
2 answers:
love history [14]4 years ago
8 0

Answer:

100 is grace's initial amount. w is how much of the substance is being decrased each week. in ryan's equation, 1 is 100% of the substance, so the initial amount, and 0.4 is the amount that is being decreased.

Step-by-step explanation:

Leona [35]4 years ago
4 0

Answer:

100= Initial Amount

1/4= decay factor for each week

w= number of weeks

1/4w= decay factor after w weeks

1 - 0.4= decay factor for each week

w= number of weeks

0.4= percent decrease

Step-by-step explanation:

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A normal distribution has a mean of 14 and a standard deviation of 4. What percent of values are from 14 to 22?
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Step-by-step explain

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3 years ago
In Applied Life Data Analysis (Wiley, 1982), Wayne Nelson presents the breakdown time of an insulating fluid between electrodes
ale4655 [162]

Answer:

The sample mean is \bar{x}=14.371 min.

The sample standard deviation is \sigma = 18.889 min.

Step-by-step explanation:

We have the following data set:

\begin{array}{cccccccc}0.15&0.82&0.81&1.44&2.70&3.28&4.00&4.70\\4.96&6.49&7.25&8.03&8.40&12.15&31.89&32.47\\33.79&36.80&72.92&&&&&\end{array}

The mean of a data set is commonly known as the average. You find the mean by taking the sum of all the data values and dividing that sum by the total number of data values.

The formula for the mean of a sample is

\bar{x} = \frac{{\sum}x}{n}

where, n is the number of values in the data set.

\bar{x}=\frac{0.15+0.82+0.81+1.44+2.7+3.28+4+4.7+4.96+6.49+7.25+8.03+8.4+12.15+31.89+32.47+33.79+36.80+72.92}{19}\\\\\bar{x}=14.371

The standard deviation measures how close the set of data is to the mean value of the data set. If data set have high standard deviation than the values are spread out very much. If data set have small standard deviation the data points are very close to the mean.

To find standard deviation we use the following formula

\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }

The mean of a sample is  \bar{x}=14.371.

Create the below table.

Find the sum of numbers in the last column to get.

\sum{\left(x_i - \overline{X}\right)^2} = 6422.0982

\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 6422.0982 }{ 19 - 1} } \approx 18.889

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3 years ago
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