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Andre45 [30]
3 years ago
10

The question is in the pic please be specific

Mathematics
1 answer:
vichka [17]3 years ago
8 0
Box 1) (LxW) 20x6=120
box 2) (LxW) 15x4=60
box 1 cost) (size of box x price of box) 120x1.25=150
box 2 cost) (size of box x price of box) 60x1.25=75
subtract 150 and 75 to get 75

answer: the company is saving $75 by choosing to make 50 of box 2 instead of 50 of box 1

hope this makes sense comment if you need more explanation
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Area of the following kite?
MakcuM [25]

Answer:

i think A=20

A=pq/2 =4·10/2 =20

Hope i helped

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6 0
4 years ago
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If a car drove 180 miles in 3 hours and a truck drove 150 miles in 2 hours, what is the speed (in miles per hour) of the faster
grigory [225]

Step-by-step explanation:

Divide 180 by 3 = 60

Divide 150 by 2 = 75

The faster vehicle is truck and its speed is 75 mph

6 0
3 years ago
Solve for x given the equation x-5+7=11
Alexeev081 [22]

Answer:

9

Step-by-step explanation:

x-5+7=11

First think of what +7 = 11 then subtract by 5

check  :)

7 0
4 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
Anyone who answers first will get brainliest
oksian1 [2.3K]

Answer: D - 4 1/2 FT.

Step-by-step explanation:

36 (length of rope) divided by 8 (sections) equals 4.5

4 0
3 years ago
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