(a) If you toss 10 coins, what percent of the time will you get the three most likely macrostates (6 heads and 4 tails, 5 heads and 5 tails, 4 heads and 6 tails)?(b) You can realistically toss 10 coins and count the number of heads and tails about twice a minute. At that rate, how long will it take on average to get either 10 heads and 0 tails or 0 heads and 10 tails?

This question is incomplete, the microstates table is left out which will be uploaded along this answer;

a) the required is 65.6%

b) the required time is 4.2666 hrs

Step-by-step explanation:

using the microstates from the table in the diagram;

a)

the percent times the above any of the above microstates occurs is;

% (6 heads and 4 tails, 5 heads and 5 tails, 4 heads and 6 tails)

=   ((210 + 252 + 210)/1024) × 100

= 0.6562 × 100

= 65.6%

Therefore the required is 65.6%

b)

the probability of getting 10 heads and 0 tail or 0 tail head and 10 tails is;

p( 10H, 0T or 0H, 10T = 2/1024

= 1/512

so on average, it will take 512 trails or at a rate of 2 trials/min, 256 mim in hours;

(256 min) ( 1h/60min) = 4.2666 hrs

Therefore the required time is 4.2666 hrs

18a

step-by-step explanation:

simply because it is. it's the highest common factor

d 192

step-by-step explanation:

we find our area to find area we multiply thoses numbers to get our answer so our area is 192