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Arada [10]
3 years ago
8

Compare 6.23x10^14 and 8.912x10^12

Mathematics
1 answer:
Tju [1.3M]3 years ago
3 0

Answer:

hello : 6.23x10^14 >  8.912x10^12^12

Step-by-step explanation:

let : A= 6.23x10^14 and  B =  8.912x10^12

A= (6.23x10^2) x10^12  ... because : 10^14 = 10^2x10^12

A = 623 x 10^12

623 > 8.912

so : A > B

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Answer:

Step-by-step explanation:

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y - y_{1} = m( x - x_{1} )

y = mx + b

~~~~~~~~~~~

(4, 5)

( - 6, 15)

m = (15 - 5) / ( - 6 - 4 ) = - 1

y - 5 = ( - 1)( x - 4 )

y =<em> (- 1) </em>x + <em>9</em>

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Len [333]

Answer:

The side s has a length of 4 and side q has a length of 4\sqrt{3} ⇒ F

Step-by-step explanation:

In the 30°-60°-90° triangle, there is a ratio between its sides

side opp (30°) : side opp (60°) : hypotenuse

       1               :         \sqrt{3}            :        2

In the given triangle

∵ The side opposite to 30° is s

∵ The side opposite to 60° is q

∵ The hypotenuse is 8

→ Use the ratio above to find the lengths of s and q

side opp (30°) : side opp (60°) : hypotenuse

       1               :         \sqrt{3}            :        2

       s               :          q             :        8

→ By using cross multiplication

∵ s × 2 = 1 × 8

∴ 2s = 8

→ Divide both sides by 2

∴ s = 4

∴ The length of s is 4

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7 0
3 years ago
Consider the two data sets below:
alina1380 [7]

Answer:

<u><em>Option c) The data sets will have the same values of their interquartile range.</em></u>

<u><em></em></u>

Explanation:

<u>1. The values are in order: </u>they are in increasing oder, from lowest to highest value.

<u>2. Calculate the interquartile range.</u>

<em />

<em>Interquartile range</em>, IQR, is the third quartile, Q3, less the first quartile Q1:

  • IQR = Q3 - Q1

To find the first and the third quartile, first find the median:

<u>Data Set 1</u>: 19, 25, 35, 38, 41, 49, 50, 52, 59

             [19, 25, 35, 38],  41,  [49, 50, 52, 59]

                                         ↑

                                     median = 41

   

<u>Data Set 2</u>: 19, 25, 35, 38, 41, 49, 50, 52, 99

             [19, 25, 35, 38] , 41,  [49, 50, 52, 99]

                                         ↑

                                      median = 41

Now find the median of each subset: the values below the median and the values above the median.

Data set 1: <u>First quartile</u>

                [19, 25, 35, 38],

                            ↑

                           Q1 = [25 + 35] / 2 = 30

                   <u>Third quartile</u>

                   [49, 50, 52, 59]

                                ↑

                                Q3 = [50 + 52] / 2 = 51

                     IQR = Q3 - Q1 = 51 - 30 = 21

Data set 2: <u> First quartile</u>

                   [19, 25, 35, 38]

                               ↑

                               Q1 = [25 + 35] / 2= 30

                  <u>Third quartile</u>

                   [49, 50, 52, 99]

                                ↑

                                Q3 = [52 + 50]/2 = 51

                   IQR = 51 - 30 = 21

Thus, it is shown that the data sets have will have the same values for the interquartile range: IQR = 21. (option c)

This happens because replacing one extreme value (in this case the maximum value) by other extreme value does not affect the median.

<em>An outlier will change the range</em> because the range is the maximum value less the minimum value.

5 0
3 years ago
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