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lilavasa [31]
2 years ago
12

Juanita receives her paycheck and knows that her gross pay and federal tax are correct. Using the fact that Social Security tax

is 6.2% of gross pay, Medicare tax is 1.45% of gross pay and state tax is 19% of federal tax, determine if Juanita's net pay is correct.
Mathematics
2 answers:
natita [175]2 years ago
8 0

Answer: letter A the net pay is correct

Step-by-step explanation:

I just took the test

bagirrra123 [75]2 years ago
5 0

Answer:

The net pay is correct.

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what is the distance around a triangle that has sides measuring 2 1/8 feet, 3 1/2 feet, and 2 1/2 feet?
Vlad1618 [11]
7 and 1/8. all you need to do is add these together by getting a common denominator of eight
5 0
2 years ago
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In △XYZ, m∠Z = 34, x = 61 cm, and z = 42 cm. Find m∠X. Round your answer to the nearest tenth of a degree.
Elza [17]
M∠X = 54.3°.

Using the Law of Sines, we have:
\frac{\sin{Z}}{z}=\frac{\sin{X}}{x}
\\
\\\frac{\sin{34}}{42}=\frac{\sin{X}}{61}

Cross multiplying gives us
61(sin 34) = 42(sin X)

Divide both sides by 42:
(61(sin 34))/42 = (42(sin X))/42
(61(sin 34))/42 = sin X

Take the inverse sine of both sides:
sin⁻¹((61(sin 34))/42) = sin⁻¹(sin X)
54.3 = X
8 0
2 years ago
During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str
Brilliant_brown [7]

Answer:

number of successes

                 k  =  235

number of failure

                 y  = 265

The   criteria are met    

A

    The sample proportion is  \r p  =  0.47

B

    E =4.4 \%

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from  the true population proportion will not  more than 4.4%

Ci  

   r =  0.514 = 51.4 \%

 v =  0.426 =  42.6 \%

D

   This 95% confidence interval  mean that the the chance of the true    population proportion of those that are happy to be exist within the upper   and the lower limit  is  95%

E

  Given that 50% of the population proportion  lie with the 95% confidence interval  the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

 Yes our result would support the claim because

            \frac{1}{3 } \ of  N    < \frac{1}{2}  (50\%) \ of \  N  , \ Where\ N \ is \ the \  population\ size

Step-by-step explanation:

From the question we are told that

     The sample size is  n  = 500

     The sample proportion is  \r p  =  0.47

 

Generally the number of successes is mathematical represented as

             k  =  n  *  \r p

substituting values

             k  =  500 * 0.47

            k  =  235

Generally the number of failure  is mathematical represented as

           y  =  n  *  (1 -\r p )

substituting values

           y  =  500  *  (1 - 0.47  )

           y  = 265

for approximate normality for a confidence interval  criteria to be satisfied

          np > 5  \ and  \ n(1- p ) \ >5

Given that the above is true for this survey then we can say that the criteria are met

  Given that the confidence level is  95%  then the level of confidence is mathematically evaluated as

                       \alpha  = 100 - 95

                        \alpha  = 5 \%

                        \alpha  =0.05

Next we obtain the critical value of  \frac{\alpha }{2} from the normal distribution table, the value is

                 Z_{\frac{ \alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

                E =  Z_{\frac{\alpha }{2} } *  \sqrt{ \frac{\r p (1- \r p}{n} }

substituting values

                 E =  1.96 *  \sqrt{ \frac{0.47 (1- 0.47}{500} }

                 E = 0.044

=>               E =4.4 \%

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from  the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

          \r p  - E <  p  <  \r p  + E

substituting values

        0.47 -  0.044 <  p  < 0.47 +  0.044

         0.426 <  p  < 0.514

The upper limit of the 95% confidence interval is  r =  0.514 = 51.4 \%

The lower limit of the   95% confidence interval is  v =  0.426 =  42.6 \%

This 95% confidence interval  mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit  is  95%

Given that 50% of the population proportion  lie with the 95% confidence interval  the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

            \frac{1}{3 }  < \frac{1}{2}  (50\%)

 

3 0
2 years ago
Kelly purchased 6 planters for a total of $18. She wants to purchase another 16 planters at the same unit price. How much will 1
AfilCa [17]

Answer:

Hope this helps

Step-by-step explanation:

If 6 planters= $18   then 16 planters= 16x  18 divided by 6 =$48

5 0
2 years ago
olivia has a shoebox shaped like a rectangular prism. She decorates 4 faces of the box, leaving the top and bottom without decor
Art [367]

Answer:

440 inches

Step-by-step explanation:

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8 0
2 years ago
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