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$Based on the following calculator output, determine the population standard deviation of the dataset, rounding to the nearest 1ooth if necessary. {:[" 1-Var-Stats "],[ bar(x)=106.142857143],[Sigma x=743],[Sigmax^(2)=80877],[Sx=18.3160091307],[sigma x=16.9573294008],[n=7],[minX=77],[Q_(1)=91],[Med^(2)=107],[Q_(3)=120],[maxX=131]:} Answer:$

Based on the following calculator output, determine the population standard deviation of the dataset, rounding to the nearest $1$ ooth if necessary. $\newline$ $\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=106.142857143 \\ \Sigma x=743 \\ \Sigma x^{2}=80877 \\ S x=18.3160091307 \\ \sigma x=16.9573294008 \\ n=7 \\ \operatorname{minX}=77 \\ \mathrm{Q}_{1}=91 \\ \mathrm{Med}^{2}=107 \\ \mathrm{Q}_{3}=120 \\ \max \mathrm{X}=131 \end{array}$ $\newline$ Answer:

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Find the limit at a vertical asymptote of a rational function II

Full solution

Q. Based on the following calculator output, determine the population standard deviation of the dataset, rounding to the nearest

1

ooth if necessary.

\newline

\begin{array}{l} \text { 1-Var-Stats } \\ \bar{x}=106.142857143 \\ \Sigma x=743 \\ \Sigma x^{2}=80877 \\ S x=18.3160091307 \\ \sigma x=16.9573294008 \\ n=7 \\ \operatorname{minX}=77 \\ \mathrm{Q}_{1}=91 \\ \mathrm{Med}^{2}=107 \\ \mathrm{Q}_{3}=120 \\ \max \mathrm{X}=131 \end{array}

\newline

Answer:

Identify symbol for population standard deviation: Identify the symbol for population standard deviation in the calculator output. $\newline$ The symbol for population standard deviation is $\sigma_x$ in the calculator output.
Locate value of standard deviation: Locate the value of the population standard deviation in the calculator output. $\newline$ The value given for ＂sigma x＂ is $16.9573294008$ .
Round standard deviation: Round the population standard deviation to the nearest hundredth. Rounding $16.9573294008$ to the nearest hundredth gives us $16.96$ .
Verify rounding accuracy: Verify that the rounding is done correctly. $16.9573294008$ rounded to two decimal places is indeed $16.96$ .

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