Here are the historical probabilities p(x) = P(X = x) that a fixed length of cast-iron piping will be manufactured and be found, on inspection, to have various numbers X of potentially weakening defects:x 0 1 2 3 4p(x) .45 .21 .13 .12 ,09F(x)Calculate the expected number of defects in a pipe E(X).

Step-by-step explanation:

For this case we have the probability density function given:

X          0        1         2       3        4

P(X)  0.45   0.21    0.13   0.12  0.09

The cdf of F(X) would be given by:

X          0         1         2        3     4

F(X)   0.45   0.66   0.79   0.91   1

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

And we can use the following formula in order to calculate the expected value:

And replacing we got:

x+y = 29, 5x + 2y = 100

step-by-step explanation:

i'm guessing it was a multiple first option was the correct answer

2 x t divided by n

step-by-step explanation: