one-third of p is at least -17
6 = p - 17 + 1-------------------5 = p - 17-------------------p = 22
22
To cube something is to raise to the third power. P cubed would be p^3
17 syllables in a haiku poem
If: -17 = 6p-5 Then: p = -2
p(one) = 1/6 p(not one) = 5/6 p(at least one) = 1 - p(failure) p(failure) = (4C0)(5/6)^4 = 1 x 625/1,296 = 0.48 p(at least one) = 1 - 0.48 = 0.52
Get contacts. :P
6 = p - 17 + 1-------------------5 = p - 17-------------------p = 22
17 S in a H P
At least one son who was born in 1994.
22
-4(3 - p) > 5(p + 1) => -12 + 4p > 5p + 5-17 > p, that is p < -17.
There is one s orbital and three p orbitals and five d orbitals in the third energy level.
The probability of tossing a coin 9 times and getting at least one tail is: P(9 times, at least 1 tail) = 1 - P(9 heads) = 1 - (0.50)9 = 0.9980... ≈ 99.8%
17 is a prime number
At least one cos i have :p
7/8 P(all heads) = (1/2)*(1/2)*(1/2)=(1/8). P(at least one tails) = 1-P(all heads) = 7/8. Another (more time-consuming) way to get the same result: P(all tails) = (1/2)^3=1/8 P(exactly two tails) = 3*(1/2)^3=3/8 P(exactly one tails) = 3*(1/2)^3=3/8 P(at least one tail) = P(all tails) + P(exactly two tails) + P(exactly one tails) = 7/8