(a) Data with the eight day's measurement.
Raw data: [60,58,64,64,68,50,57,82],
Sorted data: [50,57,58,60,64,64,68,82]
Sample size = 8 (even)
mean = 62.875
median = (60+64)/2 = 62
1st quartile = (57+58)/2 = 57.5
3rd quartile = (64+68)/2 = 66
IQR = 66 - 57.5 = 8.5
(b) Data without the eight day's measurement.
Raw data: [60,58,64,64,68,50,57]
Sorted data: [50,57,58,60,64,64,68]
Sample size = 7 (odd)
mean = 60.143
median = 60
1st quartile = 57
3rd quartile = 64
IQR = 64 -57 = 7
Answers:
1. The average is the same with or without the 8th day's data. FALSE
2. The median is the same with or without the 8th day's data. FALSE
3. The IQR decreases when the 8th day is included. FALSE
4. The IQR increases when the 8th day is included. TRUE
5. The median is higher when the 8th day is included. TRUE
10226321 because of the minus of the (1,5) 102374711
<span>Garreth and Elisa ran together, Luigi and Jasmine went the same distance, and Logan and Jasmine have the same displacement in the opposite direction. This is the answer which is C</span>
Answer:
Option a: \frac{m^{5} }{162n} is the equivalent expression.
Explanation:
The expression is \frac{(3m^{-2} n)^{-3}}{6mn^{-2} } where m\neq 0, n\neq 0
Let us simplify the expression, to determine which expression is equivalent from the four options.
Multiplying the powers, we get,
\frac{3^{-3}m^{6} n^{-3}}{6mn^{-2} }
Cancelling the like terms, we have,
\frac{3^{-3}m^{5} n^{-1}}{6 }
This equation can also be written as,
\frac{m^{5}}{3^{3}6 n^{1} }
Multiplying the terms in denominator, we have,
\frac{m^{5} }{162n}
Thus, the expression which is equivalent to \frac{(3m^{-2} n)^{-3}}{6mn^{-2} } is \frac{m^{5} }{162n}
Hence, Option a is the correct answer. 23
Step-by-step explanation:
With cross multiplication you can find that they are the same. In both equations, x would be 100.
