A sample of size \( n=10 \) is drawn from a population. The data is shown below. \( \begin{array}{lllll}106.8 & 67.2 & 85.4 & 106.8 & 95 \\ 76.7 & 75.8 & 89 & 82.2 & 106.8\end{array} \) What is the range of this data set? range \( =\square \) What is the standard deviation of this data set? (Remember, it is a sample.) Please report the answer with appropriate rounding, reporting 2 more decimal places than the original data. Please, please, please do not calculate the value by hand.

b. \( \frac{2}{3}+\frac{1}{5}=\square \square \)

Directions: Determine if segments \( A B \) and \( C D \) 1. \( \overline{A B} \) formed by \( (-2,13) \) and \( (0,3) \) \[ \begin{array}{l}\overline{C D} \text { formed by }(-5,0) \text { and }(10,3) \\ 10+5 \text { is } \frac{3}{5} 3-13 \\ 10+2\end{array} \]

\begin{tabular}{l} Compute. \\ \( \qquad \begin{array}{r}39.4+(-7.2)=32.2 \\ 6.7-39.31=\square\end{array} \) \\ \hline\end{tabular}

Compute. \[ \begin{array}{r}-7.34+39.79=\square \\ -45.3+(-71.8)=\square\end{array} \]

1. Resolve, simplifica y rodeá el resultado correcto. \( \begin{array}{llll}\text { a. } \frac{6}{7}+\frac{2}{9}: \frac{1}{3} \cdot 4= & \frac{74}{21} & \frac{128}{21} & \frac{8}{21} \\ \text { b. } \frac{8}{11} \cdot \frac{11}{3} \cdot \frac{10}{6}: \frac{7}{9}= & \frac{21}{11} & \frac{11}{21} & \frac{21}{7} \\ \text { c. } \frac{10}{9}+\frac{2}{3} \cdot \frac{13}{4}-6 \cdot \frac{1}{3}= & \frac{52}{9} & \frac{34}{9} & \frac{23}{18} \\ \text { d. }\left(\frac{1}{2}+\frac{5}{6}\right) \cdot \frac{4}{5}+\frac{1}{5}= & \frac{19}{5} & \frac{28}{5} & \frac{21}{5}\end{array} \)