2x^3y+18xy-10x^2y-90y
= 2xy(x^2 + 9)- 10y(x^2 + 9)
= (x^2 + 9)(2xy - 10y)
= (x^2 + 9) 2y(x - 5)
= 2y (x^2 + 9) (x - 5)
<span>280
I'm assuming that this question is badly formatted and that the actual number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts. So let's take each course by itself.
You can choose 1 of 7 appetizers. So we have
n = 7
After that, you chose an entre, so the number of possible meals to this point is
n = 7 * 10 = 70
Finally, you finish off with a dessert, so the number of meals is:
n = 70 * 4 = 280
Therefore the number of possible meals you can have is 280.
Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is
n = 77 * 1010 * 44 = 3421880
But I believe that it's highly unlikely that the numbers in this problem are correct. Just imagine the amount of time it would take for someone to read a menu with over a thousand entres in it. And working in that kitchen would be an absolute nightmare.</span>
Answer:
Option 3 is right
z37 is between 2 and 3 standard deviations of the mean.
Step-by-step explanation:
Let X be a random variable which represents the mean number of miles that the employees in a department live from work
X is normal (N(29,3.6)
WE have to find Z score for X
Z =
=2.22
i.e. 37 is 2.22 std deviations from the mean.
In other words, z37 is between 2 and 3 standard deviations of the mean.
