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F 100 tires are randomly selected for shipment to an outlet what is the probability that they are all good?

sergeinik [125]

<span>Lets calculate an example: Say, .001% of tires that come from the factory are bad. There is a 1/1000 chance that for any given tire randomly selected from the warehouse that a defect will be present. Each tire is a mutually exclusive independently occurring event in this case. The probability that a single tire will be good or bad, does not depend on how many tires are shipped in proportion to this known .001% (or 1/1000) defect rate. To get the probability in a case like this, that all tires are good in a shipment of 100, with a factory defect rate of .001%, first divide 999/1000. We know that .999% of tires are good. Since 1/1000 is bad, 999/1000 are good. Now, multiply .999 x .999 x .999..etc until you account for every tire in the group of 100 shipped. (.999 to the hundredth power) This gives us 0.90479214711 which rounds to about .90. or a 90% probability. So for this example, in a shipment of 100 tires, with a .001% factory defect rate, the probability is about 90 percent that all tires will be good. Remember, the tires are mutually exclusive and independent of each other when using something like a factory defect rate to calculate the probability that a shipment will be good.</span>

4 0

3 years ago

Factor completely 3x4 − 48.

kvasek [131]

Answer:

-36

Step-by-step explanation:

take

3

x

4 = 12

then - 48

=-36

8 0

3 years ago

Read 2 more answers

A = 4 0 0 1 3 0 −2 3 −1 Find the characteristic polynomial for the matrix A. (Write your answer in terms of λ.) Find the real ei

Illusion [34]

Answer:

Step-by-step explanation:

We are given the matrix

$A = \left[\begin{matrix}4&0&0 \\ 1&3&0 \\-2&3&-1 \end{matrix}\right]$

a) To find the characteristic polynomial we calculate $\text{det}(A-\lambda I)=0$ where I is the identity matrix of appropiate size. in this case the characteristic polynomial is

$\left|\begin{matrix}4-\lambda&0&0 \\ 1&3-\lambda&0 \\-2&3&-1-\lambda \end{matrix}\right|=0$

Since this matrix is upper triangular, its determinant is the multiplication of the diagonal entries, that is

$(4-\lambda)(3-\lambda)(-1-\lambda)=(\lambda-4)(\lambda-3)(\lambda+1)=0$

which is the characteristic polynomial of A.

b) To find the eigenvalues of A, we find the roots of the characteristic polynomials. In this case they are $\lambda=4,3,-1$

c) To find the base associated to the eigenvalue lambda, we replace the value of lambda in the expression $A-\lambda I$ and solve the system $(A-\lambda I)x =0$ by finding a base for its solution space. We will show this process for one value of lambda and give the solution for the other cases.

Consider $\lambda = 4$ . We get the matrix

$\left[\begin{matrix}0&0&0 \\ 1&-1&0 \\-2&3&-5 \end{matrix}\right]$

The second line gives us the equation x-y =0. Which implies that x=y. The third line gives us the equation -2x+3y-5z=0. Since x=y, it becomes y-5z =0. This implies that y = 5z. So, combining this equations, the solution of the homogeneus system is given by

$(x,y,z) = (5z,5z,z) = z(5,5,1)$