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Nady [450]
3 years ago
10

One possible solution to a diminishing Social Security payroll is to increase the Social Security tax by 1.89%. How would such a

n increase effect the tax on an annual salary of $54,000?
a.
Annual tax would increase by $1,020.60.
b.
Annual tax would increase by $2,327.40.
c.
Annual tax would increase by $3,348.00.
d.
Annual tax would increase by $4,368.60.
Mathematics
2 answers:
trasher [3.6K]3 years ago
8 0
The answer would be A).
Multiply $54,000 by 0.0189 to find out how much the tax would be increased by.
You get 1020.60
weqwewe [10]3 years ago
6 0

Answer:  Option "A" is correct.

Step-by-step explanation:

Since we have given that

Annual salary = $54000

Tax of Social Security = 1.89%

Since there is only one possible solution to a diminishing Social Security payroll is to increase the Social Security tax by 1.89%.

Annual tax would increase by

\frac{1.89}{100}\times 54000\\\\=\frac{1.89\times 54000}{100}\\\\=\$1020.6

Hence, Option "A" is correct.

You might be interested in
2748 mile flight 6 hours
Karo-lina-s [1.5K]
We Know, speed = Distance / Time
Here, Distance = 2748 miles
Time = 6 hours

Substitute their values in to the expression,
Speed = 2748 / 6 = 458 Miles / Hour

So, the speed would be: 458 miles/hour

Hope this helps!
7 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
Dennis sister charges 550 per hour before 9 p.m. for babysitting and $8 per hour after 9 p.m. she finished babysitting at 11 p.m
LiRa [457]
Since we know what time Dennis' sister finished babysitting, we can calculate how much money she's made since 9p.m. Between 9 and 11 is 2 hours, so we know she made $16 ($8 per hour x 2 hours).

That means she made $22 before 9p.m. ($32 total - $16 made past 9p.m.)

So, you just need to figure out how many hours $22 is worth and subtract that number of hours from 9p.m. That will get you to the starting point of babysitting.

$22 / $5.50 = 4     so we know that $22 is 4 hours' worth of babysitting.

9p.m. - 4 hours is 5p.m., so we know that Dennis' sister started babysitting at 5p.m. 

Hope this helps! If you liked this answer please rate it as brainliest!! Thank you!!!

3 0
3 years ago
The band sold $498 worth of tickets for their spring concert. Adult tickets cost $5 each and student tickets cost $3 each. If th
saw5 [17]
Let: A = adult tickets; S = student tickets

a. 498 = 5A + 3S
    118 = A + S

b. S = 118 - A
    498 = 5A + 3(118 - A)
    498 = 5A + 354 - 3A
    498 - 354 = 2A
    144 = 2A
     A = 72
     
    118 = S + A
    118 = 72 + S
    118 - 72 = S
    S = 46

c. Therefore, there were 72 adult tickets and 46 student tickets sold. 

5 0
4 years ago
Please need help from a math whiz
lesantik [10]
Answer-
Its 51 inches.
8 0
3 years ago
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