A VCR manufacturer receives 70% of his parts from factory F1 and the rest from factory F2. Suppose that 3% of the output from F1 are defective while only 2% of the output from F2 are defective. a. What is the probability that a received part is defective? b. If a randomly chosen part is defective, what is the probability it came from factory F1?

Step-by-step explanation:

Lets take 1000 output in total.

From these 1000 outputs, 700 are from F1 and 300 are from F2.

Defective from F1 is and defective from F2 is .

a.

Total received part is 1000.

Total defectives are (21+6) = 27.

The probability of a received part being defective is .

b.

The probability of the randomly chosen defective part from F1 is .

answer: 12

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}