In a certain city, the hourly wage of workers on temporary employment contracts is normally distributed. the mean is $15 and the standard deviation is $3. what percentage of temporary workers earn less than $12 per hour?

15.87%

Step-by-step explanation:

We have been given that in a certain city, the hourly wage of workers on temporary employment contracts is normally distributed. The mean is $15 and the standard deviation is $3.

Let us find the z-score of 12, using z-score formula.

, where,

= z-score,

= Random sample score,

= Mean,

= Standard deviation.

Upon substituting our given values in z-score formula we will get,

Let us find P(z<-1) using normal distribution table.

Therefore, probability of a temporary worker earns less than $12 per hour is 0.15866. Let us convert our given probability in percentage by multiplying by 100.

Therefore, 15.87% of temporary workers earn less than $12 per hour.

B. 16%

Step-by-step explanation: I got this right on Edmentum

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2