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mario62 [17]
3 years ago
10

What’s the perimeter can you show me how to do it

Mathematics
1 answer:
goldenfox [79]3 years ago
7 0

Answer:

48 sq. ft.

Step-by-step explanation:

P=12 times 12 which equals 144

A=12+12+12+12=48

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Write an inequality for the sentence: twice a number, decreased by the quotient of that number and 2, is at least 12
Rudiy27
For the answer to the question above,  If you've noticed, the statement mentioned "a number". Let's assign x to be the number. The first statement is twice a number (2x). Then, decreased by the quotient of that number and 2 (-x/2). Lastly, the statement at least 12 is written as >12. 
So the answer is
2x - (x/2) >12
3 0
3 years ago
Two trains, Train A and Train B weigh a total of 349 tons. Train A is heavier than Train B. The difference in their weights is 2
Ainat [17]

Answer:

Train A = 299 tons Train B = 50 tons

Step-by-step explanation:
Train A = x

Train B = y

x+y = 349

x= y +249

Plug in one equation into another

y+249+y= 349

Simplify

2y+249=349

Subtract 249 from both sides

2y=100

Divide both sides by 2

y=50

Plug it into the x equation to solve for x

x+50=349

-50 from both sides

x=299

3 0
3 years ago
Question in picture below
Artyom0805 [142]

Answer:

C

Step-by-step explanation:

7 0
3 years ago
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The data sets show the years of the coins in two collections. Derek's collection: 1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910
KATRIN_1 [288]

Answer:

Derek's collection :

Mean= 1929

Median= 1930

Range= 54

IQR = 48

MAD= 23.75

Paul's collection:

Mean= 1929

Median= 1929.5

Range= 15

IQR = 6

MAD= 3.5

Step-by-step explanation:

1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910

Mean is given by:

(1950+1952+ 1908+1902+1955+1954+1901+1910)/8

=1929

absolute deviation from mean is:

|1950-1929|= 21

|1952-1929|= 23

|1908-1929|= 21

|1902-1929|= 27

|1955-1929|= 26

|1954-1929|= 25

|1901-1929|= 28

|1910-1929|= 19

from the mean of absolute deviation gives the MAD of the data i.e.

(21+23+21+27+26+25+28+`9)/8

23.75

 

:arrange the given data to get the range and median

   1901   1902    1908   1910    1950  1952    1954   1955

The minimum value is: 1901

Maximum value is: 1955

Range is: Maximum value-minimum value

         Range=1955-1901

Range= 54

median is (1910+1950)/2

1930

   the lower set of data=

  1901   1902    1908   1910

first quartile becomes

1902+1908)/2

Q1=1905

and upper set of data is:

1950  1952    1954   1955

we find the median of the  upper quartile or third quartile is:

1952+1954)/2=1953

Q3-Q1=1953-1905=

IQR=48

 

Paul's collection:

1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920

Mean is given by:

1929+1935+ 1928+ 1930+ 1925+ 1932+1933+1920)/8

1929

absolute deviation from mean is:

|1929-1929|=0

|1935-1929|= 6

|1928-1929|= 1

|1930-1929|= 1

|1925-1929|= 4

|1932-1929|= 3

|1933-1929|= 4

|1920-1929|= 9

Hence, we get:

MAD=0+6+1+1+4+3+4+9/8

28/8

3.5

arrange the data in ascending order we get:

1920   1925   1928   1929   1930   1932   1933   1935  

Minimum value= 1920

Maximum value= 1935

Range=  15 (  1935-1920=15 )

The median is between 1929 and 1930

Hence, Median= 1929.5

Also, lower set of data is:

1920   1925   1928   1929  

the first quartile or upper quartile is

1925+1928/2

1926.5

and the upper set of data is:

1930   1932   1933   1935  

We have

1932+1933)/2

1932.5

IQR is calculated as:

Q3-Q1

6

7 0
4 years ago
does anyone have a brainly plus account I can borrow? I have a lot of classes and I don't have access to a debit/credit card.​
Mariulka [41]

Step-by-step explanation:

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5 0
3 years ago
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