The probability of rolling a die and getting a 6 is of course 1/6 and the probability of not getting a 6 is 5/6. if you roll the same die 8 times, what is the probability that 6 will show up twice?

0.261

Step-by-step explanation:

We use the Bernoulli trial method to solve this. The form of the answer goes thus:

P = nCr * x^2 * y^6

Where n is the total number of trials which is 8, while r is the expected number of success, 2 in this case. x is the probability of success 1/6 while y is the probability of failure 5/6 in this case.

P= 8C2 * (1/6)^2 * (5/6)^6

P= 28 * 0.028 * 0.335 = 0.261

answer:

the position of equilibrium will shift to the left.

step-by-step explanation:

the general equation for the dissociation of a weak base is

b + h₂o   ⇌   bh⁺ + oh⁻

according to le châtelier's principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

if you apply a stress by adding more bh⁺ or oh⁻,   the system will respond by moving in a direction that gets rid of the added ions.

the position of equilibrium will shift to the left, forming more of the undissociated base.

52 \43 i don't know this i tried