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swat32
3 years ago
12

Rewrite each number sentence using numerals and symbols. Find the difference between 4 and 3 and then add 7 *

Mathematics
1 answer:
Zina [86]3 years ago
8 0

Answer:

8

Step-by-step explanation:

4-3+7=8

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If clearer picture i may help for it is far to dark and clouded to read

4 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
3 years ago
What is answer to -14r-19=303
zubka84 [21]
-14r = 303 + 19
-14r = 322
r = -322/14

r = -23

\boxed{\huge{\sf{r=-23}}}
4 0
3 years ago
How old am I if 100 increased by 4 times my age is 132?
Setler [38]
33 years old would be correct
5 0
3 years ago
NEED HELP ASAP
Jobisdone [24]

ANSWER: A. 46

 

 

SOLUTION

 

Given that Q is equidistant from the sides of TSR

m∠TSQ = m ∠QSR

 

To solve for x

m∠TSQ = 3x + 2

m ∠QSR = 8x – 33

 

Since m∠TSQ = m ∠QSR

3x + 2 = 8x – 33

 

Add 33 to both sides

3x + 2 + 33 = 8x – 33 + 33

3x + 35 = 8x

8x = 3x + 35

 

Subtract 3x from both sides

8x – 3x = 3x – 3x + 35

5x = 35

 

Divide both sides by 5

x = 7

 

Since m∠TSQ = 3x + 2, and x = 7

m∠TSQ = (3*7) + 2

m∠TSQ = 21 + 2

m∠TSQ = 23

 

To solve for RST

Given that Q is equidistant from the sides of RST

m∠RST = m∠TSQ + m ∠QSR

Since m∠TSQ = m ∠QSR

m∠RST = 2m∠TSQ = 2m ∠QSR

 

Ginen, m∠RST = 2m∠TSQ

m∠TSQ = 23

m∠RST = 2(23)

m∠RST = 46

7 0
3 years ago
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