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vredina [299]
3 years ago
13

Your total FICA contribution which includes Social Security and Medicare is 15.3% of your salary. 12.4% of your FICA contributio

n is for Social Security. Your annual income is $67,525. What will be the total deduction be for you and your employer for Medicare?
Mathematics
1 answer:
Allisa [31]3 years ago
6 0

Answer:

$1,958.23

Step-by-step explanation:

Annual income = $67,525

Total FCIA = 15.3% of salary

Since my annual income is $67,525 the FCIA contribution = 15.3%.

But 12.4% is for Social Security.

The remaining 2.9%(15.3% - 12.4%) will be for Medicare taxes.

The total deduction for I and my employer for medicare taxes will be:

2.9% * $67,525

= $1,958.225

The total deduction for medicare would be: $1,958.23

Note: Assuming we were asked to calculate only employee's medicare, the deduction would be 50% of $1,958.23.

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Step-by-step explanation: I’m pretty sure

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2 years ago
"find the reduction formula for the integral" sin^n(18x)
dexar [7]
Let

I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx
For demonstration on how to tackle this sort of problem, I'll only work through the case where n is odd. We can write

\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx
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For the remaining integral, we can integrate by parts, taking

u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax

\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx

For this next integral, we rewrite the integrand

\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax
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So putting everything together, we found

I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx
I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)
I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax
\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax

\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax
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2 years ago
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pochemuha

SA = πrs + π r^2

subtract π r^2 from each side

SA -π r^2 =πrs

divide each side by πr

(SA -π r^2)/πr = s

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