We can use the binomial theorem to find the probability that 0 out of the 15 samples will be defective, given that 20% are defective.
P(0/15) = (15C0) (0.2)^0 (1 - 0.2)^15 = (1)(1)(0.8)^15 = 0.0352
Then the probability that at least 1 is defective is equal to 1 - 0.0352 = 0.9648. This means there is a 96.48% chance that at least 1 of the 15 samples will be found defective. This is probably sufficient, though it depends on her significance level. If the usual 95% is used, then this is enough.
(-8) - 3y= 25
-8-3y= 25
move -8 to the other side
sign changes from -8 to +8
-8+8-3y=25+8
-3y= 33
divide both sides by -3 to get y by itself
and to get +y
-3y/-3= 33/-3
Answer:
y= -11
Assuming the order required is as n-> inf.
As n->inf, o(log(n+1)) -> o(log(n)) since the 1 is insignificant compared with n.
We can similarly drop the "1" as n-> inf, the expression becomes log(n^2+1) ->
log(n^2)=2log(n) which is still o(log(n)).
So yes, both are o(log(n)).
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