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balu736 [363]
3 years ago
14

For number 1-4 find the interest. All rates are annual interests rates. Thanks

Mathematics
1 answer:
aleksandrvk [35]3 years ago
7 0
For the first one, you just multiply 0.05 x 400, which equals 20, so B is your answer.

For the second one, you multiply 0.085 x 1000, which gives you 85, but it says 3 years, so you multiply it to 3, which would be A. 255.

For the 3rd one, you multiply 0.09 x 200, which gives you 18, but it says half a year, so it would be A. 9.

For the last one, multiply 0.12 to 20,000, which is 2400, but it was 3 months, and 3 months is a quarter of a year, so multiply 0.25 x 2400. 0.25 x 2400 = 600, so your answer is B. 600.

Hope this helped you. :)
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That graph would be non-linear
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A rectangle has dimensions of 15 centimeters long and 16 meters wide what is the area of the rectangle
Brrunno [24]
240 is the answer to your question just multiply them

6 0
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Solve for x: −5|x + 1| = 10
11Alexandr11 [23.1K]

Answer:

x=10

Step-by-step explanation:

6 0
3 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
4 years ago
H(x)=3x^2−12x+16 what does h equal
Verizon [17]

Answer:

(x)=3x^2−12x+16

Answer- =3x2−12x+16

Step-by-step explanation:

Hope this helps :)

7 0
3 years ago
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