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ZanzabumX [31]
3 years ago
11

An artist is trying to choose 5 covers for children’s books. There are 10 different covers to choose from. How many ways can the

artist choose covers? (It’s a permutation and combination kind of problem)
Mathematics
1 answer:
Aleksandr [31]3 years ago
5 0

Answer:

252

Step-by-step explanation:

The order of the books isn't important, so we'll use combinations.

The number of ways to choose 5 books from 10 is:

₁₀C₅ = 10! / (5! (10 − 5)!)

₁₀C₅ = 10! / (5! 5!)

₁₀C₅ = 10×9×8×7×6 / (5×4×3×2×1)

₁₀C₅ = 252

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Fourteen hours. The equation is 5x+20=90. Solve for X.
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The area of a circle is 80.86cm2.<br> Find the length of the radius rounded to 2 DP.
weeeeeb [17]

Answer:

Solution given:

The area of a circle =80.86cm²

we have:

The area of a circle =πr²

substituting value of area of circle we get

80.86cm=3.14*r²

dividing both side by 3.14

80.86/3.14=3.14*r²/3.14

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\sqrt{25.75}=\sqrt{r²}

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Which expression has an equivalent value to x2 + 9x + 8 for all values of x?
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Day Care Tuition A random sample of 50 four-year-olds attending day care centers provided a yearly tuition average of $3987 and
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Answer:

We have to find the 90% confidence interval for the mean

                 Given sample size  

                        Sample mean  

Population standard deviation  

Here the confidence level is 90% then  

                                                  And  

The 90% confidence interval is  

Here  is the critical value at 0.05

From the tables  

Now the 90% confidence interval is =

                                                                                     =

                                                                                     =(3840.44, 4133.56)

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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

6

7 0
3 years ago
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