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ollegr [7]
3 years ago
14

A rectangular garden has dimensions of 19 feet by 16 feet. A gravel path of equal width is to be built around the garden. How wi

de can the path be if there is enough gravel for 246 square feet?
OA 5ft
OB. 3 ft
OC. 41
OD. 5.5 ft
Mathematics
1 answer:
Virty [35]3 years ago
6 0

Answer:

A rectangular garden has dimensions of 19 feet by 16 feet. A gravel path of equal width is to be built around the garden. How wide can the path be if there is enough gravel for 246 square feet?

OA 5ft

OB. 3 ft

OC. 41

OD. 5.5 ft

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What is 2 = y/5-8 plzzz answer!!!
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It's a simple linear equation in 'y'.  There is only one number that 'y' can be
that will make the equation a true statement.  That number is called the
'solution' to the equation.   Here's one way to find it:

You said that                         <u> 2  =  y/5 - 8</u>

Add  8  to each side:          10  =  y/5

Multiply each side by  5:  <em>  50  =  y</em>
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3 years ago
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Currently, the sum of the ages of Yumi, Rana and Victoria is 42 years. Four years ago, the sum of the ages of Rana and Victoria
Setler [38]
Y + R + V = 42
but 4 YEARS AGO : (R-4) +(V-4) = Y

Replace Y in the 1st equation by:  (R-4) +(V-4)

(R-4) +(V-4) + R + V = 42

2R - 2V - 8 = 42;  2R -2V  = 50, or R+V=25
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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.

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Is this shape a polygon?
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