Answer:
Step-by-step explanation:
Let 
. We have that 
if and only if we can find scalars 
such that 
. This can be translated to the following equations:
1. 
2.
3. 
Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for 
and check if the third equationd is fulfilled.
Case (2,6,6)
Using equations 1 and 2 we get
whose unique solutions are 
, but note that for this values, the third equation doesn't hold (3+2 = 5 
6). So this vector is not in the generated space of u and v.
Case (-9,-2,5)
Using equations 1 and 2 we get
whose unique solutions are 
. Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.
the probability that if you pick only one, it's defective, is 250/6700
Therefore, the probability that one is not defective is 6450/6700
a. You want all 4 to not be defective: (6450/6700)^4
b. all 100 have to be not defective: (6450/6700)^100
If you type this into a calculator, you will get about 0.022, so a probability of 2 % that all of them are not defective. As this is a very small probability, the outlet should plan with returned tires.
The solution to this is 3(8)-3(9).
Answer:
v
Step-by-stedcvp explanation: welcome to family fued
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