the mean cost of domestic airfares is RS 385 per ticket. Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of RS 110. What is the probability that a domestic airfare is RS 550 or more?

0.0668

Step-by-step explanation:

Given the following :

Mean (m) Cost = RS385

Standard deviation (s) = RS110

Assume a normal distribution, Probability that domestic airfare is RS550 or more?

P(x > 550)

Find the z - score

Z - score = (x - m) / s

Where x = 550

Z = (550 - 385) / 110

Z = 165 / 110 = 1.5

P(Z > 1.5) = 1 - P(Z ≤1.5)

Using the z table : 1.5 = 0.9332

Therefore,

1 - P(Z ≤1.5) = 1 - 0.9332 = 0.0668

P(x > 550) = 0.0668

b is 38 degrees.

a is 52 degrees.

this is because if b is 38 and a is 14 degrees more than b than 38+14=52 and 52+38=90 degrees. hope this .

answer: about 22.67

step-by-step explanation: