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.
The answer is
C) -7
Multiply -5 by 2
Divide 8 by 4 :
3-10+2-2
Subtract 10 from 3 :
-7+2-2
Add -7 and 2 :
-5 - 2
Subtract 2 from -5:
-7
Start with 1.
1st number = 1
2nd number = 1 * 4 + 3 = 7
3rd number = 7 * 4 + 3 = 31
4th = 31 * 4 + 3 = 127
5th = 127 * 4 + 3 = 511
Which means the 6th = 511 * 4 + 3, which is 2047.
