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4 years ago

3 2/3 as a decimal Please help need to now

Mathematics

2 answers:

Ratling [72]4 years ago

5 0

2/3= 0.6

3.06 would be the answer I believe.

Hope this Helps!!

lina2011 [118]4 years ago

4 0

3.666 is the answer to your question

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Vicki ate 15 cookies last week. If she ate 5 cookies Friday, what fraction of the week's cookies did she eat Friday? What fracti

Anika [276]

Vicki ate 5/20 cookies on Friday and 15/20 last week

7 0

2 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

7 0

4 years ago

The Chicago Cubs have played in 28 less than twice the number of World Series Champion Than the St. Louis Cardinals. The cubs pl

aev [14]

The answer to your question is 11 world championships

8 0

3 years ago

Find the inverse of the function.

asambeis [7]

Answer:

A) $f^{-1}(n)= (n+2)^3$

Step-by-step explanation:

$f(n) = \sqrt[3]{n} -2$

$y = \sqrt[3]{n} -2$

$y = n^{1/3}-2$

$n = y^{1/3} -2$

$n + 2 = y^{1/3}$

$y = (n + 2) ^3$

$f^{-1}(n) = (n + 2)^3$

Hope this helps!

7 0

3 years ago

Read 2 more answers

The data for Shawn's phone has 32000 megabytes if storage for photos . She wants to increase the amount by 4% if each photo uses

Law Incorporation [45]

Answer:

Yes, she will have have enough new memory for 200 additional photos.

Step-by-step explanation:

Given:

Initial memory for storage, $M_{o}=32000\textrm{ megabytes}$

Increase in memory $=4%$

Therefore, increase in memory, $\Delta M=\frac{4}{100}\times 32000=1280\textrm{ megabytes}$

Now, memory consumed by 1 photo = 5 megabytes

So, memory consumed by 200 additional photos = $200\times 5=1000\textrm{ megabytes}$

Memory left in her phone = 1280 - 1000 = 280 megabytes

Therefore, she will have additional 280 megabytes of memory left in her phone.

As memory consumed by 200 additional photos is less than that of the increase in memory, therefore, she does have enough new memory for 200 additional photos.

6 0

3 years ago