The given operation involves just swapping the two rows, so carrying out
on the matrix gives
![\left[\begin{array}{cc|c}5&5&5\\1&-\dfrac23&5\end{array}\right]\to\left[\begin{array}{cc|c}1&-\dfrac23&5\\5&5&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D5%265%265%5C%5C1%26-%5Cdfrac23%265%5Cend%7Barray%7D%5Cright%5D%5Cto%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%26-%5Cdfrac23%265%5C%5C5%265%265%5Cend%7Barray%7D%5Cright%5D)
Answer:
Step-by-step explanation: It's 34
Assuming these are 4^(1/7), 4^(7/2), 7^(1/4) and 7^(1/2), the conversion process is pretty quick. the denominator, or bottom, of your fraction exponent becomes the "index" of your radical -- in ∛, "3" is your index, just for reference. the numerator, aka the top of the fraction exponent, becomes a power inside the radical.
4^(1/7) would become ⁷√4 .... the bottom of the fraction becomes the small number included in the radical and the 4 goes beneath the radical
in cases such as this one, where 1 is on top of the fraction radical, that number does technically go with the 4 beneath the radical--however, 4¹ = 4 itself, so there is no need to write the implied exponent.
4^(7/2) would become √(4⁷) ... the 7th power goes with the number under your radical and the "2" becomes a square root
7^(1/4) would become ⁴√7 ... like the first answer, the bottom of the fraction exponent becomes the index of the radical and 7 goes beneath the radical. again, the 1 exponent goes with the 7 beneath the radical, but 7¹ = 7
7^(1/2) would become, simply, √7
The correct answer would be D, the median 86.5.
In this set of data most of the values are in the 80's or 90's. There is only 1 number that is outside of this range, that is the value 0. It is an outlier.
When we have an outlier, the median generally gives us the better measure of central tendency.
Answer:
120
Step-by-step explanation:
to find the volume you just do length x width x height.
