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Vinvika [58]
2 years ago
12

A soccer player drank 3 liters of water during a game. How many milliliters did the soccer player drink?

Mathematics
2 answers:
Lapatulllka [165]2 years ago
8 0

Answer:

3000 mL

Step-by-step explanation:

recall 1 liter = 1000 mL

hence

3 liters = 1000 mL x 3 = 3000 mL

Brums [2.3K]2 years ago
3 0

Answer: 3000 milliliters

Step-by-step explanation:

To convert liters to milliliters , we will multiply through by 1000

Therefore :

3 liters = 3 x 1000 milliliters

3 liters = 3000 milliliters

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Many credit card companies charge a compound interest rate of 1.8% per month on a credit card balance. Nelson owes $850 on a cre
Finger [1]

The sequence will be


850, 850 \times (1.018)^1, 850 \times (1.018)^2, 850 \times (1.018)^3, 850 \times (1.018)^4, ...


That's


850, 865.30, 880.875, 896.731, 912.872, ...



8 0
3 years ago
Read 2 more answers
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
<img src="https://tex.z-dn.net/?f=what%20is%20the%20square%20root%20of%20245u%20to%20the%20power%20of%204%20v%20to%20the%20power
Effectus [21]

Answer:

hope it helps ......

Step-by-step explanation:

....

have a great day

5 0
2 years ago
Scenario:
KiRa [710]

Answer:

do you still need help with this?

Step-by-step explanation:

5 0
2 years ago
Is 7 14 28 56 arithmetic geometric neither or both
nadezda [96]

Answer:

it is a geometric sequence

Step-by-step explanation:

the sequence starts at 7 and doubles each time. a geometric sequence is a series of numbers multiplied each time buy a previous constant

3 0
3 years ago
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