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? a. 6% b. 16% c. 26% d. 36%

B. 16%

Step-by-step explanation:

In the picture attached, the standard normal distribution table is shown.

The value of Z is computed as follows:

Z = (x - μ)/σ

where x is the value of interest ( = 12), μ is the mean ( = 15) and σ is the standard deviation (= 3). Replacing:

Z = (12 - 15)/3 = -1

From the table, the probability is 0.1587, approximately equal to 16%

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

c) the median for town a 30" is greater than the median for town b 25"

step-by-step explanation:

town a, median = q2 = 30

town b, median - q2 = 25

so the median for town a 30" is greater than the median for town b 25"