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blondinia [14]
3 years ago
8

On this assignment, round all answers (when necessary) to exactly three decimal places. The numbers of construction workers for

various projects are 32 20 25 52 16 21 28 35 23 41 46 17 23 27 The mean of this data is The standard deviation of this data is The five-number summary is { , , , , } The Interquartile Range (IQR) is Are there any outliers according to the IQR rule (type 0 for yes, 1 for no) Replacing the largest number with results in the smallest whole number outlier.
Mathematics
1 answer:
german3 years ago
3 0

Answer:

Mean = 29

S.D = 10.95

NO outlier in the data          

Step-by-step explanation:

We are given the following data:

n = 14

Construction Workers: 32, 20, 25, 52, 16, 21, 28, 35, 23, 41, 46, 17, 23, 27

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}

where x_i are data points, \bar{x} is the mean and n is the number of observations.

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean = \frac{406}{14} = 29

Standard Deviation =

\sqrt{\frac{9,+ 81+16+ 529+ 169+ 64+ 1+36+ 36+ 144+ 289+ 144+ 36+ 4}{13}}\\= \sqrt{\frac{1558}{13} } = 10.94

Five number summary:

Data = 16,17,20,21,23,23,25,27,28,32,35,41,46,52

Minimum = 16

Maximum = 52

Median = Mean of 25 and 27 = 26

First Quartile = 21

Third Quartile = 35

Interquartile range = Q_3 - Q_1 = 35 - 21 = 14

Outliers:

\text{Lower limit} = Q_1 - (1.5)IQR = 21 - 21 = 0\\\text{Upper limit} = Q_3 + (1.5)IQR = 35 + 21 = 56\\

There is no outlier in the data.

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3 years ago
How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?
BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:  

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}

Where n = 8 and r = 0,1....8

When r = 0

\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1

When r = 1

\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 2

\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8

When r = 3

\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 4

\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70

When r = 5

\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 6

\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28

When r = 7

\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 8

\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0
3 years ago
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