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Trava [24]
3 years ago
14

Lisa and Kate are playing a card game, and a total of 900 points has been scored. Lisa scored 250 more points than Kate. If you

let l=l=the number of points that Lisa scored, and k=k= the number of points that Kate scored, then the problem can be represented by the system:
l+k=900l+k=900 and l=k+250l=k+250

Graph the system. How many points did each of them score?

Select one:
a. Kate = 575 and Lisa = 325
b. Kate = 250 and Lisa = 650
c. Kate = 450 and Lisa = 450
d. Kate = 325 and Lisa = 575

Mathematics
2 answers:
ludmilkaskok [199]3 years ago
8 0

Answer:

Lisa score 575 points and Kate scored 325 points.

Step-by-step explanation:

We are given the following information:

Total score for Lisa and Kate is 900 and Lisa scored 250 more points than Kate.

Scores for Lisa and Kate can be expressed with the help of two equations:

L + K = 900\\L = K + 250 \Rightarrow L-K = 250

Solving these equations we have:

L + K + L - K = 900 + 250\\2L = 1150\\L = 575\\K = 575 - 250 = 325

Hence, Lisa score 575 points and Kate scored 325 points.

The graph of the equations is attached.

Olegator [25]3 years ago
5 0
d. Kate = 325 and Lisa = 575 

because it says that LIsa has scored 250 points more.
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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
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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

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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

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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

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