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
Jesus is the lord above all
Answer:
1/3 and -1/2
Step-by-step explanation:
You can use the factoring method to identify the solutions of x
6x^2+x-1
2 numbers have to multiply to 6 and subtract to 1 because x is the same as 1x
So, you get 3x and 2x
since you know that -1x1 is -1 you will get
(3x-1)(2x+1)=0
Now, since (3x-1) and (2x+1) multiply to get 0, if either (3x-1) or (2x+1) is 0 the statement will be valid
3(1/3)-1=0
2(-1/2)+1=0
So, your answers are 1/3 and -1/2