Anormal distribution has a mean of 142 and a standard deviation of 16.
what is the probability that a randomly selected value lies between 142 and 174?

Mean = 142;
Standard Deviation ( "Sigma" ) = 16
172 = 142 + 32 = 142 + 2 * 16 = Mean + 2 Standard Deviation
It means that between 142 and 174 lies :
34 % + 13.5 % = 47.5 % of all values.
The probability that a randomly selected value lies between 142 and 174 is:
P = 0.475 

Let x be a random variable in the population.
P(142 < x < 174) = P((142 - 142)/16 < z < (174 - 142)/16) = P(0 < z < 2) = P(z < 2) - P(z < 0) = 0.97725 - 0.5 = 0.47725

The answer for you question is b

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

0 = ( 2 - 4 ) ( p + q )0 = ( 2 - 4 ) ( 2 + 5 i ),   by multiplying these0 = ( 4 + 10 i - 8 - 20 i )0 = 4 - 8 + 10 i - 20 i0 = - 4 -10 i   0 = 2 ( - 2 - 5 i )0 / 2 = - 2 - 5 i- 2 - 5 i = 0- 5 i = 2i = 2 / - 5