Answer:
395, 17.212
(360.576,429.424)
Step-by-step explanation:
Given that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p.
Here 
Since np and nq are greater than 5 by rule of thumb we can approximate binomial to normal.
Mean = np = 395
Variance = npq = 
Std dev = 17.212
Thus X no of successes is N(395, 17.212)
THe the minimum usual value mu minus 2 sigma and the maximum usual value mu plus 2 sigma would be

-4 2/5 / -5
-4 2/5 = -22/5
-5 = -5/1
-22/5 / -5/1
Flip -5/1 to multiply
-22/5 x 1/-5
Numerator x numerator / denominator x denominator
-22 x 1 = -22
5 x -5 = -25
-22/-25 = 22/25
The answer is 22/25.
Answer:
25/2 a^2 and the alternative form 12.5a^2
Step-by-step explanation:
Answer:

Step-by-step explanation:

