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VikaD [51]
3 years ago
14

Kelly takes out a loan with the stated rate of 12% interest. If the interest is calculated weekly, how much greater is Kelly’s e

ffective rate (The growth rate of compounded amount) Then her stated rate (the “stated” 12%)? Does the difference in effective rate and stated rate change depending on 1 year or 10 years ?
Please answer this step if you can! Greatly appreciate it.
Mathematics
1 answer:
xeze [42]3 years ago
3 0

Answer:

12.73%

Step-by-step explanation:

Given the annual rate is 12% compounded weekly.

Let's assume 1 yr=52weeks.

The effective interest rate is calculated as:

i_m=(1+r/m)^m-1, r=annual\  rate\\\\i_{52}=(1+0.12/52)^{52}-1\\\\i_{52}=0.1273

Hence, the effective rate for 12% compounded weekly is 12.73%

-The effective rate will be different when the number of compounding times is less than or more than 1 year.

- the effective rate is equal to the stated rate if the compounding number of times is once per year.

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3 years ago
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4 years ago
The following data set represents the age of
Julli [10]

\huge\text{Hey there!}


\huge\textsf{Breaking the meaning down in simpler terms}
\huge\textbf{Median:}\\\large\text{is understood to be \bf the number in the center (also known as}\\\large\textbf{the middle number)}


\huge\textbf{Mean:}\\\large\text{is understood to be the \bf total.}

\huge\textsf{Formulas for each term}

\huge\text{Median:}\\\large\textsf{To find the median you have to put each of your numbers in your data}\\\large\textsf{set from LEAST (smallest) to GREATEST (biggest). }


\huge\text{Mean:}\\\mathsf{\dfrac{sum\ of\ all\ terms}{number\ of\ terms}= mean}


\huge\textbf{SOLVING FOR THE QUESTION}

\huge\textsf{Median:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textbf{Conversion: }\large\text{27, 33, 38, 41, 42, 52, 64, 68, 72}\\\\\large\textsf{Make sure it is even between both sides of the data plot.}\\\\\large\text{It seems to even on BOTH sides of  \boxed{\textsf{14}} so it could possibly be your}\\\large\text{median.}

\huge\textsf{Mean:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textsf{Your equation: }\mathsf{\dfrac{27 + 52 + 64 + 41 + 33 + 38 + 42 + 60 +72 + 68}{10}}\\\\\mathsf{\dfrac{79 + 64 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{143 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{184 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{217 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{255 + 42 + 60 + 72 + 68}{10}}

\mathsf{\dfrac{297 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{357 + 72 + 68}{10}}\\\\\mathsf{\dfrac{429 + 68}{10}}\\\\\mathsf{\dfrac{497}{10}}\\\\\mathsf{\approx 49.70}\\\\\large\text{From the looks of it \boxed{\rm{\dfrac{497}{10}}}\ or \boxed{\text{49.70}} could possibly be your mean.}


\huge\text{Good luck on your assignment \& enjoy your day!}

<h3>
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8 0
3 years ago
Three students collected more than $93 washing cars. They used $15 to reimburse their parents for cleaning supplies. Then they d
Alenkasestr [34]
First step is find profit so lets subtract 93 and 15.
93-15=78
Finally we can divide profit by 3 students, so
78:3=26$ its the answer
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4 years ago
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