In a certain section of southern california, the distribution of monthly rent for a one-bedroom apartment has a mean of $2,200 and a standard deviation of $250. the distribution of the monthly rent does not follow the normal distribution. in fact, it is positively skewed. what is the probability of selecting a sample of 50 one-bedroom apartments and finding the mean to be at least $1,950 per month? (round z value to 2 decimal places and final answer to 4 decimal places.)

The distribution of  will still use a normal distribution though the distribution is positively skewed. 

Mean μ = 2200
Standard deviation σ = 250

Standard error σ / √ n = 250 / √ 50 = 35.36
standardize xbar to z = (xbar - μ) / (σ / √ n)
P(xbar > 1,950) = P( z > (1,950-2200) / 35.36)
= P(z > -7.07) = 1

So probability is 1 or virtually certain.

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!

i dont know

step-by-step explanation:

umm. i dont know