Use technology or a z-score table to answer the question.

The nightly cost of hotels in a certain city is normally distributed with a mean of $180.45 and a standard deviation of $24.02.

Approximately what percent of hotels in the city have a nightly cost of more than $200?

The percentage of hotels in the city have a nightly cost of more than $200 is 21%

Step-by-step explanation:

Since the nightly cost of hotels in a certain city is normally distributed,

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = the nightly cost of hotels.

µ = mean cost

σ = standard deviation

From the information given,

µ = $180.45

σ = $24.02

The probability that a hotel in the city has a nightly cost of more than $200 is expressed as

P(x > 200) = 1 - P(x ≤ 200)

For x = 200,

z = (200 - 180.45)/24.02 = 0.81

Looking at the normal distribution table, the probability corresponding to the z score is 0.79

Therefore,

P(x > 200) = 1 - 0.79 = 0.21

The percentage of hotels in the city have a nightly cost of more than $200 is

0.21 × 100 = 21%

2 over 5

step-by-step explanation:

the 2: 3 ratio is the length of the ac segment to the cb segment total length. this implies that the ratio between segment ac length and segment ab length is 2: 5. point c is 2/5 from a to b, so the required fraction is 2/5.

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>