Answer:
Step-by-step explanation:
8 3/5 on a number line is between 8 and nine
idk how to draw a number line on a computer so ill just tell you how
1. divided the spaces between 8 and 9 into 5 sections ( 3/5 )
2. then mark 3 of those 5 spaces and put a dot or a line at where the 3rd space ends
Answer:
The best choice would be hiring a random employee from company A
Step-by-step explanation:
<em>Supposing that the performance rating of employees follow approximately a normal distribution on both companies</em>, we are interested in finding what percentage of employees of each company have a performance rating greater than 5.5 (which is the mean of the scale), when we measure them in terms of z-scores.
In order to do that we standardize the scores of both companies with respect to the mean 5.5 of ratings
The z-value corresponding to company A is
where
= mean of company A
= 5.5 (average of rating between 1 and 10)
s = standard deviation of company A
We do the same for company C
This means that 27.49% of employees of company C have a performance rating > 5.5, whereas 71.42% of employees of company B have a performance rating > 5.5.
So, the best choice would be hiring a random employee from company A
In order to do this, you must first find the "cross product" of these vectors. To do that, we can use several methods. To simplify this first, I suggest you compute:
‹1, -1, 1› × ‹0, 1, 1›
You are interested in vectors orthogonal to the originals, which don't change when you scale them. Using 0,-1,1 is much easier than 6s and 7s.
So what methods are there to compute this? You can review them here (or presumably in your class notes or textbook):
http://en.wikipedia.org/wiki/Cross_produ...
In addition to these methods, sometimes I like to set up:
‹1, -1, 1› • ‹a, b, c› = 0
‹0, 1, 1› • ‹a, b, c› = 0
That is the dot product, and having these dot products equal zero guarantees orthogonality. You can convert that to:
a - b + c = 0
b + c = 0
This is two equations, three unknowns, so you can solve it with one free parameter:
b = -c
a = c - b = -2c
The computation, regardless of method, yields:
‹1, -1, 1› × ‹0, 1, 1› = ‹-2, -1, 1›
The above method, solving equations, works because you'd just plug in c=1 to obtain this solution. However, it is not a unit vector. There will always be two unit vectors (if you find one, then its negative will be the other of course). To find the unit vector, we need to find the magnitude of our vector:
|| ‹-2, -1, 1› || = √( (-2)² + (-1)² + (1)² ) = √( 4 + 1 + 1 ) = √6
Then we divide that vector by its magnitude to yield one solution:
‹ -2/√6 , -1/√6 , 1/√6 ›
And take the negative for the other:
‹ 2/√6 , 1/√6 , -1/√6 ›
Answer:
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Step-by-step explanation:
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