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
So the distance formula: d = rad (x2 - x1)^2 + (y2 - y1)^2
it doesn't matter what order you do, you just have to make sure it's a y coordinate for y and an x coordinate for x.
rad (9-5)^2 + ((-6)-1)^2
rad 4^2 + (-7)^2
rad 16 + 49
rad 65
i think we have to find the answer for terms of a
p=a+4b+2c
ISOLATE A
p-4b-2c=a
answer:
p-4b-2c=a
A will be 7 x 5
B will be 10x10