suppose we have a fuse box containing 40 fuses of which 6 are defectives. If two fuses are selected at random and removed from the box. Find the probability that both are defective, if the first fuse (a) Replaced (b) Not replaced.

a) P(1&2 =defect)= 1/800

b)  P(1&2 =defect)= 1/780

Step-by-step explanation:

a) The probability that 1st of the selected fuses is defective is   2/40=1/20 =0.05

So if we replace it by the not defective the number of defective fuses is 1 and total number is 40.

So the probability that 2-nd selected fuse is defective as well is 1/40

The probability both fuses are defective is

P(1&2 =defect)= 2/40*1/40=2/1600=1/800

b) The probability that 1st of the selected fuses is defective is   2/40=1/20 =0.05

SO residual amount of the fuses is 39. 1 of them is defective.

So the probability that 2-nd selected fuse is defective as well is 1/39

The probability both fuses are defective is

P(1&2 =defect)= 2/40*1/39=2/1560=1/780

my guess would be b and d, but i didn’t actually get out the pen and paper. lol

in 2010, the population of the state of california is 37.29 × 10⁶

step-by-step explanation:

population of los angeles = 3.79 × 10⁶

population of california = 9.84 × population of los angeles

                                        = 9.84 × 3.79 × 10⁶

                                        = 37.2936 × 10⁶

                                        = 37.29 × 10⁶

∴ in 2010, the population of the state of california is 37.29 × 10⁶