New York City is the most expensive city in the United States for lodging. The room rate is $204 per night (USA Today, April 30, 2012). Assume that room ratesa mally distributed with a standard deviation of $55. a. What is the probability that a hotel room costs $225 or more per ni b. What is the probability that a hotel room costs less than $140 per n c. What is the probability that a hotel room costs between $200 and $300 per ni me 25. are nor ght? ight? d. What is the cost of the most expensive 20% of hotel rooms in New York City?

a. P=0.35

b. P=0.12

c. P=0.49

d. Cost: more than $249 per night.

Step-by-step explanation:

We assume a normal distribution with mean $204 and s.d of $55.

To calculate the probability we calculate the z-value for each case

a. What is the probability that a hotel room costs $225 or more

b. What is the probability that a hotel room costs less than $140

c. What is the probability that a hotel room costs between $200 and $300

d. What is the cost of the most expensive 20% of hotel rooms in New York City?

In this case, we have to estimate z' so thats P(z>z')=0.2. This value is z=0.841.

Then, the value of X should be:

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube

the second option

step-by-step explanation: