Answer:
option 0.8%
Step-by-step explanation:
Data provided in the question:
Mean = 5.7 years
Standard deviation, s = 1.8 years
Now,
P(the employee has worked at the store for over 10 years)
= P(X > 10 years)
= 
or
= 
= P (Z > 2.389 )
or
= 0.008447 [from standard z table]
or
= 0.008447 × 100% = 0.84% ≈ 0.8%
Hence,
the correct answer is option 0.8%
Answer:
a) independent = hours
dependent = wages
b ) W = 10 + 5h
c)27.50
Step-by-step explanation:
The independent variable is the one that you change which is the number of hours that it takes
The dependent variable is the variable that changes based on the independent variable. The wages depends on the hour worked so the wages are the independent variable
The wages equal the fixed cost plus the hourly rate times the number of hours where W is the wages and h is the number of hours
W = 10 + 5h
W = 10 + 5( 3.5)
= 10 + 17.5
= 27.5
1. Add -4.5 to 18.5
2. Divide 23 by 7
3. Then subtract 6 from -2.7
12 divided by 2/3 = 18
2/3 divided by 12 = 0.5