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The IQR measures middle spread: ( \text{IQR} = Q_3 - Q_1). First, confirm the data are sorted (they are). With 9 values, the median is the 5th value → 12. The lower half is the first four values: (6, 7, 9, 10). For an even count, (Q_1) is the average of the two middle lower-half values: ((7 + 9)/2 = 8). The upper half is the last four values: (15, 18, 20, 30). (Q_3) is the average of the two middle upper-half values: ((18 + 20)/2 = 19). Therefore, ( \text{IQR} = 19 - 8 = 11). Wait—recheck arithmetic carefully: ((7 + 9) = 16), (16/2 = 8) correct; ((18 + 20) = 38), (38/2 = 19); (19 - 8 = 11). But 11 is not an answer choice, so ensure we partitioned correctly. Alternative convention (exclusive median removed, halves as 4 values each) was applied; that’s standard. Let’s validate values: Did we misread the set? (6, 7, 9, 10, 12, 15, 18, 20, 30). Yes. Some curricula use inclusive method for odd (n), including the median in halves. Using inclusive: lower half (6, 7, 9, 10, 12) → (Q_1 = 9); upper half (12, 15, 18, 20, 30) → (Q_3 = 18); IQR (= 18 - 9 = 9) (C). TEAS commonly uses the inclusive method. Thus the correct IQR is 9. Choose option C. (Note: Be aware of method specified; if unspecified on TEAS, inclusive halves for odd (n) are typical.)