suppose the people have weights that are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Find the probability that if a person is randomly selected, his weight will be greater than 174 pounds?
Assume that weights of people are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Mean = 177
standard deviation = 26
We find z-score using given mean and standard deviation
z = 
= 
=-0.11538
Probability (z>-0.11538) = 1 - 0.4562 (use normal distribution table)
= 0.5438
P(weight will be greater than 174 lb) = 0.5438
Answer:
∠PTQ = 124°
Step-by-step explanation:
∠QPT = 90° - 62° = 28°
∠PQT = 28°
∠PTQ = 180° - 28° - 28° = 124°
Given:

Solution:
Complex formula:

Let us simplify one by one.


|3 + 4i| = 5


|3 - 4i| = 5


|-3 + 4i| = 5


|-3 - 4i| = 5
Substitute these in the given expression:


The solution of the expression is 20.
To find out how much more it rained, you subtract 1 1/4 or 1.25 and 3/5.
So you get 1.25-.6, which is .65 or 13/20 inches.
Hope this helps.