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Bezzdna [24]
3 years ago
12

How many sixths are in 7 1/6

Mathematics
2 answers:
DochEvi [55]3 years ago
7 0

Answer:

43

Step-by-step explanation:

7 1/6 ÷ (1/6) = 43

43 x (1/6) = 7  1/6

Hence, the answer (43) is correct

NNADVOKAT [17]3 years ago
6 0

All we need to do is turn the mixed number into an improper fraction.

7 1/6 = 43/6

Therefore, there are 43 sixths in 7 1/6.

Best of Luck!

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The answer is:  90° .
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Help with #7 and #8?
arsen [322]

Answer:

M(6,5) and G(2,9)

Step-by-step explanation:

m = ( \frac{x1 + x2}{2}, \frac{y1 + y2}{2} )

B(3,6)

C(9,4)

m = [(3+9)/2 , (6+4)/2]

= (6,5)

M(5,9)

H(8,9)

G(x,y)

[(x+8)/2 , (y+9)/2]=(5,9)

(x+8)/2 = 5

x+8 = 10

x = 2

(y+9)/2 = 9

y+9 = 18

y = 9

G(2,9)

(Correct me if i am wrong)

5 0
3 years ago
For number 12, it says write an equation in slope-intercept for for the total cost of any number of tickets at 7 tickets for $5.
ra1l [238]

Find the y-intercept then model it into y = .71x + b (b is the y-intercept).

4 0
3 years ago
Find the number of all 20 digit integers in which no two consecutive digits are the same
Arturiano [62]
For the first digit of the required 20-digit number, we can have 9 choices for the number, those are the digits 1 to 9. In the second digit of the number, we will only have 8 choices. That is because, we don't want to write a number in the second digit that is similar to the first digit and the third digit.

Similarly, in the third digit's place, we will only have 8 choices because we do not want the digit to be similar to the second digit and the fourth digit. This will go on up until the 19th digit. In the 20th digit, we will also have 9 choices (from 0-9) but we do not want the digit to be similar to the 19th digit. 

From the explanation given above, we have the number of ways,

    n = 9 x 8^(19 - 1) x 9  = 1.459 x 10^18

<em>ANSWER: 1.459 x 10^18</em>
5 0
4 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
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