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Gemiola [76]
3 years ago
8

Write an equation of the line that passes through (-2, -4) and (-4, 2).

Mathematics
1 answer:
Effectus [21]3 years ago
5 0
You can formulate an equation of the line given two points using the two-point slope form:

m = (y2-y1) / (x2-x1)

Wherein, after plugging in the values, it will now be on the form:

y = mx + b

I hope my answer has come to your help. God bless and have a nice day ahead!
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PLSSSSS HELPPP
lorasvet [3.4K]
Do 2.63 x 3 to check ur answer
3 0
3 years ago
Could someone explain to me what is the process
Phoenix [80]

Answer:

You just multiple the number of apples one truck can carry by the number of trucks total.

The second one you divide 4551 by 37 and round your answer up.

Step-by-step explanation:

So it would be 23*37.

3 0
3 years ago
Read 2 more answers
Evaluate each expression by replacing the variable with 8. Show each step. (8 marks) 2n - 6
monitta

Answer:

10

Step-by-step explanation:

2n-6

n=8

2(8)-6

16-6

10

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
3 years ago
What's the prime factor of 92 ?
mariarad [96]
Prime factors are multiplying numbers that cannot be divided into smaller numbers.
Let's break down 92.
Divide 92 by 2.
2 x 46.
We can still break this down.
Divide 46 by 2.
2 x 2 x 23.
This is your prime factorization of 92.
I hope this helps!
8 0
3 years ago
Read 2 more answers
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