Iknow the answer but i need to understand how to get it. bailey will roll a number cube and flip a coin for a probability experiment. the faces of the cube are labeled 1-6. the coin can land on heads or tails. if bailey rolls the number cube once and flips the coin once which list contains only the outcomes in which the number cube lands on a number greater than 4.
the right answer was a list with says:
5, heads
5, tails
6, heads
6, tails

Total output= H5, H6,T5, T6

Step-by-step explanation:

Bailey will roll a number cube and flip a coin for a probability experiment.

The faces of the cube are labeled 1-6= 6 possible outcomes

The coin can land on heads or tails= 2 possible outcomes

If Bailey rolls the number cube once= 1 output

and flips the coin once = 1 output

Total output will be combined outcome of cube and coin:

       possible outcome of cube= 1, 2, 3 , 4 , 5, 6

       possible outcome of coin= H, T

                                    (H for heads and T for tails)

Total output =possible combined outcome of cube and coin= H1, H2, H3 , H4 , H5, H6, T1, T2, T3 , T4 , T5, T6

Now

the outcomes in which the number cube lands on a number greater than :

reduces the total required output by removing the combination when the number cube lands on a number less than or equals to 4

therefore removing H1, H2, H3 , H4 and T1, T2, T3 , T4 from the total outcome will result in following list:

Total output= H5, H6,T5, T6 i.e.

5, Heads

5, Tails

6, Heads

6, Tails !

2

step-by-step explanation:

1+1=2

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.