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blsea [12.9K]
3 years ago
5

Hayden has 6 rolls of dimes there are 50 dimes in each roll. How many dimes does he have altogether

Mathematics
2 answers:
Anestetic [448]3 years ago
8 0

Answer:

Hayden has total 300 dimes altogether.

Step-by-step explanation:

Hayden has 6 rolls of dimes.

There are 50 dimes in each roll.

Multiply 50 dimes by 6 to get number of total dimes

Total dimes he have altogether = 50 × 6 = 300 dimes.

Hayden has total 300 dimes altogether.

DerKrebs [107]3 years ago
3 0
This is pretty simple. :) Since you have 6 rolls of dimes with 50 dimes in each, you would just do 6*50 which equals 300. So Hayden has 300 dimes altogether. Hope this helps! :)
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D = b^2 -4ac solve for a 2/3 (x-18y) = 6y solve for x
seropon [69]

Answer:

a = \frac{b^2-d}{4c}

x = 27y

Step-by-step explanation:

Order of Operations: BPEMDAS

<u>Question 1</u>

Step 1: Write out discriminant

d = b² - 4ac

Step 2: Subtract b² on both sides

d - b² = -4ac

Step 3: Divide both sides by -1

b² - d = 4ac

Step 4: Divide both sides by 4c

a = \frac{b^2-d}{4c}

<u>Question 2</u>

Step 1: Write out equation

2/3(x - 18y) = 6y

Step 2: Distribute parenthesis

2/3x - 12y = 6y

Step 3: Isolate <em>x </em>by adding 12y on both sides

2/3x = 18y

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x = 27y

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2 years ago
Juan's income y consists of at least $37,500 salary plus 5% commission on all of his sales x. which inequality represents juan's
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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
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

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3 years ago
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maw [93]

Answer:

C. 2x + 6

Step-by-step explanation:

After doing synthetic division (see attached), you get 2, 6, and 0, corresponding to the answer 2x + 6.

5 0
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