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Ber [7]
3 years ago
13

Answer please. Need help I'm giving a follow, a thanks, and a brainliest.

Mathematics
2 answers:
Misha Larkins [42]3 years ago
6 0

mean  =add them all up and divide by the number of points

(17.2+18.3+14.9+16.8+15.3)/5

82.5/5

16.5

Choice B

GalinKa [24]3 years ago
3 0

16.5

Add the data set together and then divide by 5, or how many numbers are in the data set

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Each investment matures in 3 years. The interest compounds annually.
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bearing in mind that 4¾ is simply 4.75.

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$600\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=600\left(1+\frac{0.05}{1}\right)^{1\cdot 3}\implies A=600(1.05)^3\implies A=694.575 \\\\[-0.35em] ~\dotfill

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$750\\ r=rate\to 4.75\%\to \frac{4.75}{100}\dotfill &0.0475\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=750\left(1+\frac{0.0475}{1}\right)^{1\cdot 3}\implies A=750(1.0475)^3\implies A\approx 862.032

well, the interest for each is simply A - P

695.575 - 600 = 95.575.

862.032 - 750 = 112.032.

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