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OleMash [197]
3 years ago
12

5 feet 4 feet What is the perimeter? 16 feet 14 feet 18 feet

Mathematics
1 answer:
brilliants [131]3 years ago
8 0

perimiter means all the way around, and im assuming this is a 4-sided shape, so 5+5+4+4 = 18 ft.

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Express the ratio 7day to 6weeks as a decimal fraction<br>​
Whitepunk [10]

Answer:

6 weeks=6*7 days=42days

7/42 =1/6 =0.16667

OR

7 days=1 week

therefore 1/6=0.16667

Note that both must be in the same unit.

8 0
3 years ago
Find The Area Of This Parallelogram. Be Sure To Include The Correct Unit In Your Answer. (10 Points Guaranteed)
schepotkina [342]

Answer:

132ft²

Step-by-step explanation:

area of a parallelogram = base length × height

base length = 11 ft and height = 12ft

so area = 11 × 12 = 132ft²

5 0
3 years ago
The given lengths are two sides of a right triangle. All three side lengths of the triangle are integers and together form a Pyt
Ludmilka [50]
12 & 35 Third side is hypotenuse = 37
28 & 45 Third side is hypotenuse = 53
20 & 29 Third side is a leg = 21

4 0
3 years ago
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
2x + 3y = 3 and x + 6y = -3
jekas [21]

Answer:

Step-by-step explanation:

Multiplying x+6y=-3 with 2 so its equal to first eqn,

Eliminating,

2x + 3y = 3        (1)

2x +12y = -6       (2)                  (because subtracting)

------------------

      -9y =9

     ∴y=-1

eq. y in 1

2x -3=3

2x=3+3

x=6/2

=3

Pls mark me brainliest

4 0
3 years ago
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