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Sauron [17]
4 years ago
7

115.9 rounded to the nearest hundred

Mathematics
2 answers:
dybincka [34]4 years ago
8 0

Answer:

100

Step-by-step explanation:

115.9 is closer to 100 then it is to 200, so round down to 100

weqwewe [10]4 years ago
7 0

Answer:

100

Step-by-step explanation:

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At an interest rate of 8% compounded annually, how long will it take to double the following investments?
Paladinen [302]
Let's see, if the first one has a Principal of $50, when it doubles the accumulated amount will then be $100,

recall your logarithm rules for an exponential,

\bf \textit{Logarithm of exponentials}\\\\
log_{{  a}}\left( x^{{  b}} \right)\implies {{  b}}\cdot  log_{{  a}}(x)\\\\
-------------------------------\\\\
\qquad \textit{Compound Interest Earned Amount}
\\\\


\bf A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$100\\
P=\textit{original amount deposited}\to &\$50\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}
\\\\\\
100=50\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 100=50(1.08)^t
\\\\\\
\cfrac{100}{50}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\


\bf log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\
-------------------------------\\\\


now, for the second amount, if the Principal is 500, the accumulated amount is 1000 when doubled,

\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$1000\\
P=\textit{original amount deposited}\to &\$500\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}
\\\\\\
1000=500\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 1000=500(1.08)^t
\\\\\\


\bf \cfrac{1000}{500}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\
-------------------------------

now, for the last, Principal is 1700, amount is then 3400,

\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$3400\\
P=\textit{original amount deposited}\to &\$1700\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}

\bf 3400=1700\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 3400=1700(1.08)^t
\\\\\\
\cfrac{3400}{1700}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t
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