Answer:
Step-by-step explanation:
![6k^3+10k^2-56k\\2k(3k^2+5k-28)\\=2k[3k^2+12k-7k-28]\\=2k[3k(k+4)-7(k+4)]\\=2k(k+4)(3k-7)](https://tex.z-dn.net/?f=6k%5E3%2B10k%5E2-56k%5C%5C2k%283k%5E2%2B5k-28%29%5C%5C%3D2k%5B3k%5E2%2B12k-7k-28%5D%5C%5C%3D2k%5B3k%28k%2B4%29-7%28k%2B4%29%5D%5C%5C%3D2k%28k%2B4%29%283k-7%29)
common factor is 3k-7
Zero.
Anything times zero is zero.
-20 * -10 = 200
200 * -1 = -200
-200 * 0 = 0
<span>Quartiles are generally more reliable for judging outliers than mean and standard deviations for 2 reasons. The mean is simply the average of all of the numbers, meaning that an outlier can easily be obscured by the masses. Standard deviation is a better method, however only going over by one standard deviation in either direction would also mask an outlier. A strong outlier however will pull a quartile farther in that direction than would normally be expected.</span>
The rate per hour is equal to the total bagels made divided by the time to make them.
360 ÷ 10 = 36
If for 1 hour they make 36 bagels, for 13 hours they can bake
36 x 13 = 468 bagels.
