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Ipatiy [6.2K]
2 years ago
7

Module Quiz: Modified

Mathematics
1 answer:
Minchanka [31]2 years ago
8 0

Answer:

b I think.

Step-by-step explanation:

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Using Order of Operations (PEMDAS), solve the following two expressions. Show all steps.
aksik [14]

Answer:

A:

Multiple: 3 * 5 = 15

Add: 7 + the result of step No. 1 = 7 + 15 = 22

Divide: 4 / 2 = 2

Subtract: the result of step No. 2 - the result of step No. 3 = 22 - 2 = 20

Add: the result of step No. 4 + 3 = 20 + 3 = 23

B:

Add: 7 + 3 = 10

Multiple: the result of step No. 1 * 5 = 10 * 5 = 50

Divide: 4 / 2 = 2

Subtract: the result of step No. 2 - the result of step No. 3 = 50 - 2 = 48

Add: the result of step No. 4 + 3 = 48 + 3 = 51

8 0
2 years ago
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938 divided by 7, pls give the solution too​
stellarik [79]

Answer:

134

Step-by-step explanation:

938÷7=134

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2 years ago
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Type the correct answer in each box. Use numerals instead of words. Any non-integer answers in this problem should be entered as
ladessa [460]

Answer:

The interquartile range of the data set is 5

The mean absolute deviation of the data set is 3.6

Step-by-step explanation:

* Lets explain how to find the interquartile range and the mean absolute

  deviation (MAD)

- The steps to find the interquartile range is:

1- Arrange the values from the smallest to the largest

∴ The values are 5 , 6 , 10 , 11 , 12 , 13 , 14 , 15 , 18 , 20

2- Find the median

- The median is the middle value after arrange them

* If there are two values in the middle take their average

∵ The values are 10 then the 5th and the 6th are the values

∵ The 5th is 12 and the 6th is 13

∴ The median = \frac{12+13}{2}=12.5

∴ The median is 12.5

3- Calculate the median of the lower quartile

- The lower quartile is the median of the first half data values

∵ There are 10 values

∴ The first half is the first five values

∴ The first half values are 5 , 6 , 10 , 11 , 12

∵ The middle value is 10

∴ The median of lower quartile = 10

- Similar find the median of the upper quartile

- The upper quartile is the median of the second half data values

∵ There are 10 numbers

∴ The second half is the last five values

∴ The second half values are 13 , 14 , 15 , 18 , 20

∵ The middle value is 15

∴ The median of upper quartile = 15

4- The interquartile range (IQR) is the difference between the upper

    and the lower medians

∴ The interquartile range = 15 - 10 = 5

* The interquartile range of the data set is 5

* Lets talk about the mean absolute deviation

- Mean absolute deviation (MAD) of a data set is the average distance  

 between each data value and the mean

- To find the mean absolute deviation of the data, start by finding

   the mean of the data set.  

1- Find the sum of the data values, and divide the sum by the  

   number of data values.  

∵ The data set is 5 , 6 , 10 , 11 , 12 , 13 , 14 , 15 , 18 , 20

∵ Its sum = 5 + 6 + 10 + 11 + 12 + 13 + 14 + 15 + 18 + 20 = 124

∵ The mean = the sum of the data values/the number of the data

∵ The set has 10 numbers

∴ The mean = 124/10 = 12.4

2- Find the absolute value of the difference between each data value  

   and the mean ⇒ |data value – mean|

# I5 - 12.4I = 7.4

# I6 - 12.4I = 6.4

# I10 - 12.4I = 2.4

# I11 - 12.4I = 1.4

# I12 - 12.4I = 0.4

# I13 - 12.4I = 0.6

# I14 - 12.4I = 1.6

# I15 - 12.4I = 2.6

# I18 - 12.4I = 5.6

# I20 - 12.4I = 7.6

3- Find the sum of the absolute values of the differences.

∵ Their sum = 7.4 + 6.4 + 2.4 + 1.4 + 0.4 + 0.6 + 1.6 + 2.6 + 5.6 + 7.6

∴ Their sum = 36

4- Divide the sum of the absolute values of the differences by the

    number of data values  to find MAD

∴ MAD = The sum of the absolute values/number of the values

∵ The sum = 36

∵ The data set has 10 values

∴ MAD = 36/10 = 3.6

* The mean absolute deviation of the data set is 3.6

6 0
2 years ago
How many different ways can you make change for $.50 using only nickels,dimes, and quarters?
yanalaym [24]

Answer:

10 ways

Step-by-step explanation:


5 0
3 years ago
Q16. Find x^2 + 1/x^2 , if x + 1/x = 3
erica [24]

Answer:

2 \times  x \times \frac{1}{x}  =  2 \\  {x}^{2}   +  {y }^{2}  + 2xy =  {(x + y)}^{2} \\ {x}^{2}  +  \frac{1}{ {x}^{2} }  + 2 = {(x +  \frac{1}{x} })^{2}  =  {3}^{2}  = 9

8 0
2 years ago
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