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Doss [256]
3 years ago
15

Tax returns filled manually have a 20% chance of containing errors while tax returns filled electronically have a .05% chance of

containing the same if 2.7 million tax returns are filed each way how many more erroneous manually filed returns will there be than erroneous electronically filed returns
Mathematics
2 answers:
Travka [436]3 years ago
7 0

Answer:

538,650

Step-by-step explanation:

We must first find how many errors there will be if filed manually and if filed electronically

Manually: 2,700,000*20% or 2,700,000*.2

Answer: 540,000 errors

Electronically: 2,700,000*.05% or 2,700,000*.0005

Answer: 1,350 errors

We must then find the difference; 540,000-1,350=538,650

Gelneren [198K]3 years ago
4 0

Answer:

538,650.

Step-by-step explanation:

Number of erroneous manual returns = 20% of 2.7 million

= 0.20 * 2,700,00 = 540,000.

Number for electronically returned =  2,700,000 * 0.0005 = 1350.

Difference = 540,00 - 1350 = 538,650.

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\bf \qquad \textit{Amount for Exponential Growth}\\\\
A=I(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\stackrel{multiplier}{\textit{initial amount}}\\
r=\stackrel{increase~rate}{rate\to r\%\to \frac{r}{100}}\\
t=\textit{elapsed time}\\
\end{cases}\\\\
-------------------------------

\bf \textit{we know that in }\stackrel{year~0}{2000}\textit{ there were }\stackrel{A}{\$25}\implies 25=I(1+r)^0
\\\\\\
25=I(1)\implies 25=I\qquad \qquad \boxed{A=25(1+r)^t}\\\\
-------------------------------\\\\
\textit{we also know that in }\stackrel{year~10}{2010}\textit{ it turned to }\stackrel{A}{72}\implies 72=25(1+r)^{10}

\bf \cfrac{72}{25}=(1+r)^{10}\implies \sqrt[10]{\frac{72}{25}}=1+r\implies \sqrt[10]{\frac{72}{25}}-1=r
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6 0
3 years ago
Which expression is a fourth root of -1+isqrt3?
aleksklad [387]

Answer:

Step-by-step explanation:

\sf n^{th} roots of a complex number is given by DeMoivre's formula.

   \sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}

Here, k lies between 0 and (n -1) ; n is the exponent.

\sf -1 + i\sqrt{3}

a = -1 and b = √3

\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}

\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}

                   \sf \theta = tan^{-1} \ \dfrac{\sqrt{3}}{-1}\\\\ = tan^{-1} \ (-\sqrt{3})

                   \sf = \dfrac{-\pi }{3}

n = 4

For k = 0,

          \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin  \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi  }{12}+iSin  \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]

For k =1,

         \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]

For k =2,

       z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]

For k = 3,

      \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]

For k = 4,

      \sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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