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Arte-miy333 [17]
3 years ago
10

A bag contains colored tiles.

Mathematics
2 answers:
olga2289 [7]3 years ago
5 0

Answer:

50%

Step-by-step explanation:

there are 12 total, half are green

earnstyle [38]3 years ago
5 0

Answer:

Hello There!!

Step-by-step explanation:

the tiles of green are 6 and the total tiles are 12 so its 6/12=1/2=0.5

hope this helps,have a great day!!

~Pinky~

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\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the first 3 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$20000\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases}

\bf A=20000\left(1+\frac{0.06}{1}\right)^{1\cdot 3}\implies A=20000(1.06)^3\implies \boxed{A=2382.032} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the next 4 years}}

\bf A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2382.032\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}

\bf A=2382.032\left(1+\frac{0.08}{1}\right)^{1\cdot 4}\implies A=2382.032(1.08)^4\implies \boxed{A\approx 3240.73} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{amount for this period}}{2382.032+3240.73}\implies 5622.762

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