Two types of defects are observed in the production of integrated circuit boards. It is estimated that 6% of the boards have solder defects and 3% have some surface-finish defects. The occurrences of the two types of defects are assumed to be independent of each other. If a circuit board is randomly selected from the output, find the probabilities for the following situations:(a) Either a solder defect or a surface-finish defect or both are found.(b) Only a solder defect is found.(c) Both types of defects are found.(d) The board is free of defects.(e) If a surface-finish

(a) 0.09

(b) 0.06

(c) 0.0018

(d) 0.9118

(e) 0.03

Step-by-step explanation:

Let A = boards have solder defects and B = boards have surface defects.

The proportion of boards having solder defects is, P (A) = 0.06.

The proportion of boards having surface-finish defects is, P (B) = 0.03.

It is provided that the events A and B are independent, i.e.

(a)

Compute the probability that either a solder defect or a surface-finish defect or both are found as follows:

= P (A or B) + P (A and B)

Thus, the probability that either a solder defect or a surface-finish defect or both are found is 0.09.

(b)

The probability that a solder defect is found is 0.06.

(c)

The probability that both defect are found is:

Thus, the probability that both defect are found is 0.0018.

(d)

The probability that none of the defect is found is:

Thus, the probability that none of the defect is found is 0.9118.

(e)

The probability that the defect found is a surface finish is 0.03.

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

e

step-by-step explanation:

an equation by definition has to have two or more variables (x and 2 in this case)