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anonymous
 one year ago
The Mendes family bought a new house 11 years ago for $100,000. The house is now worth $196,000. Assuming a steady rate of growth, what was the yearly rate of appreciation? Round your answer to the nearest tenth of a percent (1.2% etc)
anonymous
 one year ago
The Mendes family bought a new house 11 years ago for $100,000. The house is now worth $196,000. Assuming a steady rate of growth, what was the yearly rate of appreciation? Round your answer to the nearest tenth of a percent (1.2% etc)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0growth : y = a(1+r)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[196000 = 100000 (1 + \frac{ r }{ 100 } ) ^{11}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would i multiplu 100 to both sides ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0divide by 100000 on both sides

phi
 one year ago
Best ResponseYou've already chosen the best response.1you probably should be using the continuous interest formula http://cs.selu.edu/~rbyrd/math/continuous/ \[ P= P_0 e^{r \ t} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[1.96 = (1+\frac{ r }{ 100 })^{11}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now can i divide by 100

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no you need to take log

phi
 one year ago
Best ResponseYou've already chosen the best response.1if you assume compounding at a yearly rate, your answer will be off by a few tenths of a percent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea it asked for yearly rate of appreciation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0log1.96=log (1+r/100)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how can there be power after you log

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\log_{11} (1 + \frac{ r }{ 100 }) = 1.96 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\log_{11} (1.96) = 0.28064\]

phi
 one year ago
Best ResponseYou've already chosen the best response.1***Assuming a steady rate of growth*** means continuous they want the annual interest rate assuming continuous compounding.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea thats what we've been using

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay i about to solve it tell me if im correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I believe you are to assume annual compounding. By "steady rate of growth" I believe the author simply means that the annual rate of appreciation is constant over the duration of the problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dayum if thats the case its simple interest formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you move the decimal two places forward or back

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And base 11 logs are not required. You have\[1.96=\left( 1+\frac{ r }{ 100 } \right)^{11}\]Using base 10 logs\[\log1.96 = 11 \log \left( 1+\frac{ r }{ 100 } \right)\]\[\frac{ \log 1.96 }{ 11 } = \log \left( 1+\frac{ r }{ 100 } \right)\]\[0.02657 = \log \left( 1+\frac{ r }{ 100 } \right)\]\[1+\frac{ r }{ 100 } = 10^{0.02657} = 1.0631\]Can you take it from here and solve for \(r\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no i dont understand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would it be .0631/100 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So close. \[1+\frac{ r }{ 100 } = 0.0631\]Subtract 1 from both sides\[\frac{ r }{ 100 } = 0.0631\]How do you solve for r?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry. First line should read\[1+\frac{ r }{ 100 } = 1.0631\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You are correct. Now round to the number of decimal places required. Well done?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now you need to round off somemore

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can yall help me with another ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Delete this question and post the other.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it said that was wrong . it says to round to the nearest tenth of percent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i told you to round off somemore lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it still says its wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We both indicated that you were to round to the correct number of decimal places :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its still wrong . i put 6.3%

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well. I don't know what the issue is. \[196000 = 100000\times 1.603^{11}\]so the math checks out.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wanna try simple interest method?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can't imagine that it's continuous compounding but, if so, you'll have to use natural logs.\[F=Pe^{rt}\]\[196000 = 100000e^{11r}\]\[1.96 = e^{11r}\]\[\ln 1.96 = 11r\]\[r=\frac{ \ln 1.96 }{ 11 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Convert to percent by multiplying by 100% and round to the correct number of decimal places.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hate that kinda englishtesting maths questions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Me too. Just finished a graduate course in financial mathematics and "steady rate of growth" was never used to indicate continuous compounding. @phi , if this is the correct answer, I owe you an apology. Please forgive my doubt.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorta can't understand how e comes into play in continuous compounding

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0THANK YALL !!!! . can yall help me with the next one
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