]Eigenvectors are found by the equation

implying that

. We then can write:
And:
Gives us the characteristic polynomial:

So, solving for each eigenvector subspace:
![\left [ \begin{array}{cc} 4 & 2 \\ 5 & 1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} -x \\ -y \end{array} \right ]](https://tex.z-dn.net/?f=%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%204%20%26%202%20%5C%5C%205%20%26%201%20%5Cend%7Barray%7D%20%5Cright%20%5D%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bc%7D%20x%20%5C%5C%20y%20%5Cend%7Barray%7D%20%5Cright%20%5D%20%3D%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bc%7D%20-x%20%5C%5C%20-y%20%5Cend%7Barray%7D%20%5Cright%20%5D%20)
Gives us the system of equations:
Producing the subspace along the line

We can see then that 3 is the answer.
Answer:
-2( 7x+4)
Step-by-step explanation:
-14x - 8
Factor out -2
-2*7x -2*4
-2( 7x+4)
Answer:
$45.50
Step-by-step explanation:
1. $65.00 ÷ 100 = 0.65
2. 0.65 x 30 = $19.50
3. $65.00 - $19.50 = $45.50
Answer:
There is some inforrmation that is missing in this question. It should read:
A container holds 50 electronic components, of which 10 are defective. If 6 components are drawn at random from the container, the probability that at least 4 are not defective is . If 8 components are drawn at random from the container, the probability that exactly 3 of them are defective is .
Answers
Part 1. 0.02
Part 2. 0.0375
Explanation
The probability is a chance of an event happening. It is calculated as;
probability = (Number of favourable outcome)/(Number of available outcome)
Part 1
6 are chosen at random. If 4 are not defective, then 2 are defective.
P(at least 4 are not defective) = 4/40 × 2/10
= 1/10 ×1/5
= 1/50
= 0.02
Part 2
8 are chosen at random. If 3 are defective, the 5 are not defective.
P(3 are defective) = 3/40 × 5/10
= 15/400
= 3/80
= 0.0375
Step-by-step explanation: