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

5x + 5

step-by-step explanation:

you should always add the diagrams and graphs we cannot see. this particular question is kind of borderline because we can make a graph for you.

(f + g)x = -x^2 +3x + 5 + x^2 + 2x

(f + g)x = 5x + 5

the graph of z = 5x + 5 is shown below. always use desmos when in doubt.