In a certain section of Southern California, the distribution of monthly rent for a one-bedroom apartment has a mean of $2,075 and a standard deviation of $300. 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 55 one-bedroom apartments and finding the mean to be at least $1,985 per month

Probability is 1

Step-by-step explanation:

We are given;

mean;μ = $2,075

Standard deviation;σ = $300

n = 55

x' = $1,985

Now, we want to find x' to be at least $1,985 which is P(x' > $1,985).

The z-value is calculated from;

z = (x' - μ)/(√σ/n)

Plugging in the relevant values;

z = (1985 - 2075)/(√300/55)

z = -38.536

So, P(x' > $1,985) = P(z > -38.536)

This transforms to;

P(z < 38.536)

Probability from z distribution table is 1

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)

believe it or not anna is correct good job anna give me a fist bump there u go gal