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Anarel [89]
4 years ago
9

34.9 was rounded to 1 decimal place. What is the upper bound?

Mathematics
1 answer:
Luba_88 [7]4 years ago
8 0
<h3>Answer:  34.95</h3>

=======================================================

Explanation:

The upper boundary, or upper bound, is a fence of sorts.

Note how the following values all round to 34.9, when rounding to one decimal place

  • 34.94
  • 34.945
  • 34.949
  • 34.9499
  • 34.94999
  • 34.949999
  • 34.9499999
  • 34.94999999

We can keep going with the pattern of 9s at the end and whatever value you form will still round to 34.9; so you can see that there is no largest value we can reach that will round to 34.9

All of those values are slowly approaching 34.95, which is the upper fence I mentioned. We can get closer and closer to this value, but we can't actually reach it. Otherwise, 34.95 will round to 35.0

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A bicycle wheel with radius 26" rotates through an arc that measures 80°. What is the length of the arc of the tire that touched
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What is the area of the parallelogram?
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<h2>486mm²</h2>

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4 0
2 years ago
Calculate the standard deviation of the following data. 3, 4, 5, 6, 2, 3, 12, 79, 5
malfutka [58]
The answer to this is 23.42 (For population, SD)

The data given as a whole is called UNGROUPED data. The standard deviation can be computed with the formula: 

\sqrt{\dfrac{\sum(x_i-\bar{x})^2}{n}}

To do this, you need to first figure out the following:

Mean (\bar{X}) =?
The sum of \sum(x_i-\bar{X})^{2} =?

You need to write down your data in a table, but before you can do this, you need to first find out what the mean is and you can do this by adding up all the data and dividing it by the number of observations:

\bar{X} = \frac{sum of all data}{number of observations}
\bar{X} = \frac{3+4+5+6+2+3+12+79+5}{9} = \frac{119}{9}= 13.22

So we subtract the mean from each data to fill in the second column. I'll use the first raw data to make the first row as an example here:
(x-\bar{X}= 3-13.22=-10.22) 

For the third column, you just need get the square of the second column.
(x_i-\bar{X})^{2} =(-10.22)^{2}= 104.4484

For this table, let x be raw data and X be \bar{X}:

  x   |    (x-X)        |    (x-X)²     
  3   |    -10.22     |   104.4484         
       |                   |
  4   |   -  9.22      |     85.0084
       |                   |
  5   |   -  8.22      |     67.5684
       |                   |
  6   |   -  7.22      |     52.1284
       |                   |
  2   |     - 11.22   |     125.8884
       |                   |
  3   |    -10.22     |     104.4484
       |                   |
12   |   -  1.22      |         1.4884 
       |                   |
79   |    65.78      |   4327.0084
       |                   |
 5    |    -  8.22     |       67.5684
                 
REMEMBER: The Σ means "the sum of" to get Σ(x-X)² Add the values in the third column to get the sum of (x-X)² and you will get 4,935.5556 

Now all you have to do is input it into your equation:

sd=\sqrt{\dfrac{\sum(x_i-\bar{x})^2}{n}}
\sqrt{\frac{4935.5556}{9}}
\sqrt{548.39506666}
23.42

The standard deviation is 23.42

4 0
4 years ago
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