Abatch consists of 12 defective coils and 88 good ones. find the probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made. 0.7744 0.7733 0.0144 0.176

The probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made is 0.7744  

Solution:

Total number of coils = number of good coils + defective coils = 88 + 12 = 100

p(getting two good coils for two selection) = p( getting 2 good coils for first selection ) p(getting 2 good coils for second selection)

p(first selection) = p(second selection) =

Hence, p(getting 2 good coil for two selection) =

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~

too many. (jk)

31 v bucks

step-by-step explanation:

since you bought 25 on friday and 6 on tuesday, you can just add them up.

25+6=31

check your work.

31-6=25

i hope this !

-yand3r3