If an employee works on a holiday, he or she is typically paid
D. two times the employee's hourly rate.
This holiday pay is also referred to as double time or double the normal rate of pay because the worker is paid two times the hourly rate. but, no longer all employees qualify for vacation Pay: it depends on several aspects like collective bargaining settlement insurance, in case you work in a civil carrier position, hourly rate or if your employer presents time beyond regulation for working on a holiday.a holiday is an afternoon set apart via custom or by regulation on which normal activities, in particular business or work along with faculty, are suspended or reduced.
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Solution:
Weight of person = 179 lbs
As given person needs IV Drip in m L.
Also,→ 10 mg = 2 m L
Now, you must remember , → 1 lbs = 0.453 Kg (approx)
→179 lbs = 179 × 0.453 Kg = 78.47 Kg (approx)
The person needs at least 60 mg for each Kg they weigh.
⇒ 1 Kg → 60 mg
⇒ 78.47 × 1 Kg → 78.47 × 60 mg
⇒ 78.47 Kg → 4708.2 mg
Also, 10 mg → 2 m L
⇒ 1 mg → 
m L
⇒ 4708 . 2 mg → 
m L
⇒ 4708 . 2 mg → 941. 64 ml
When you divide two fractions, you do Keep, Change, Flip or KCF. So you keep the 3/5, you change the division sign to a multiplication sign, then flip the 3/15 to 15/3. So the new equation is 3/5x15/3. Then you simply the two sides so it becomes 1/1x3/1.....
Meaning the answer is 3.
Answer:
(x-3), 4 (x - 3)^2 (x + 3) (2 x + 7)
Step-by-step explanation:
Factor all the expressions,
1st expression= 4x^2 - 36=4(x^2-9)=4(x+3)(x-3)
2nd expression=2x^2 - 12x + 18 =2(x^2-6x+9)=2 (x - 3)^2=2(x-3)(x-3)
3rd expression=2x^2 + x - 21=(x - 3) (2 x + 7)
HCF=Commo factor=(x-3)
LCF=Common factor*Remaining factor=4(x+3)(x-3)(x-3) (2 x + 7)=4 (x - 3)^2 (x + 3) (2 x + 7)
Answer:
B. No, this distribution does not appear to be normal
Step-by-step explanation:
Hello!
To observe what shape the data takes, it is best to make a graph. For me, the best type of graph is a histogram.
The first step to take is to calculate the classmark`for each of the given temperature intervals. Each class mark will be the midpoint of each bar.
As you can see in the graphic (2nd attachment) there are no values of frequency for the interval [40-44] and the rest of the data show asymmetry skewed to the left. Just because one of the intervals doesn't have an observed frequency is enough to say that these values do not meet the requirements to have a normal distribution.
The answer is B.
I hope it helps!
