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Ann [662]
3 years ago
15

I need help with this question!!!!

Mathematics
1 answer:
Genrish500 [490]3 years ago
7 0

Answer: The answer is A over B if you know what im sayin

Step-by-step explanation:

You take A and B and put the A over the B.

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Answer:

Step-by-step explanation:

2x+y

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Answer:

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Step-by-step explanation:

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The savings account offering which of these APRs and compounding periods offers the best APY?
ANTONII [103]
A)

\bf \qquad  \qquad  \textit{Annual Yield Formula}
\\\\
\left.  \qquad \qquad\right. \left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 3.1196\%\to \frac{3.1196}{100}\to &0.031196\\
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\left.  \qquad \qquad\right. \left(1+\frac{0.031196}{2}\right)^{2}-1

b)

\bf \qquad  \qquad  \textit{Annual Yield Formula}
\\\\
\left.  \qquad \qquad\right. \left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 3.1184\%\to \frac{3.1184}{100}\to &0.031184\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four times}
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\left.  \qquad \qquad\right. \left(1+\frac{0.031184}{4}\right)^{4}-1

c)

\bf \qquad  \qquad  \textit{Annual Yield Formula}
\\\\
\left.  \qquad \qquad\right. \left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 3.1095\%\to \frac{3.1095}{100}\to &0.031095\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{daily, assuming 365 days}
\end{array}\to &365
\end{cases}
\\\\\\
\left.  \qquad \qquad\right. \left(1+\frac{0.031095}{365}\right)^{365}-1

d)

\bf \qquad  \qquad  \textit{Annual Yield Formula}
\\\\
\left.  \qquad \qquad\right. \left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 3.1172\%\to \frac{3.1172}{100}\to &0.031172\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
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\end{array}\to &12
\end{cases}
\\\\\\
\left.  \qquad \qquad\right. \left(1+\frac{0.031172}{12}\right)^{12}-1
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Try the suggested explanation (it consists of 3 basic steps).

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