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solong [7]
3 years ago
14

Last month Grace worked, and was paid for, a total of x hours. Some of the hours were on the day shift and the remainder of the

hours were on the night shift. Her hourly pay is 20% higher for the night shift than for the day shift. How many hours did Grace work on the day shift last month
Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
7 0

Say her hourly pay was $p for the day shift and $1.2p for the night shift.

(1) x = 55. Clearly insufficient: we need to know how these 55 hours are split between the day shift hours and the night shift hour.

(2) Grace's gross pay for the hours she worked on day shift last month was exactly 50% of her total gross pay for last month. Not sufficient.

(1)+(2) Say Grace worked h hours on the day shift and 55-h hours on the night shift. Thus, her pay for the the hours she worked on day shift is $ph and the total gross pay is $ph+(55-h)*1.2p. From (2) we have that ph=0.5(ph+(55-h)*1.2p) --> reduce by p: h=0.5(h+(55-h)*1.2). We have 1 variable, so we can solve for it. Sufficient.

Answer: C.

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Describe the distribution of the data and if the mean or median would be a better measure of center for each. Include outliers.
Bumek [7]

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Is there a picture you can take

Step-by-step explanation:

8 0
3 years ago
Next number in sequence 3 -6 12 4 20 ?
Artyom0805 [142]
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8 0
3 years ago
What is the missing angle measurement in the polygon shown below?
Zinaida [17]

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6 0
3 years ago
Read 2 more answers
(4 pts) If a rock is thrown vertically upward from the surface of Mars with an initial velocity of 15m/sec
Yuliya22 [10]

Answer:

Step-by-step explanation:

I see you're in college math, so we'll solve this with calculus, since it's the easiest way anyway.

The position equation is

s(t)=-1.86t^2+15t  That equation will give us the height of the rock at ANY TIME during its travels. I could find the height at 2 seconds by plugging in a 2 for t; I could find the height at 12 seconds by plugging in a 12 for t, etc.

The first derivative of position is velocity:

v(t) = -3.72t + 15 and you stated that the rock will be at its max height when the velocity is 0, so we plug in a 0 for v(t):

0 = -3.72t + 15 and solve for t:\

-15 = -3.72t so

t = 4.03 seconds. This is how long it takes to get to its max height. Knowing that, we can plug 4.03 seconds into the position equation to find the height at 4.03 seconds:

s(4.03) = -1.86(4.03)² + 15(4.03) so

s(4.03) = 30.2 meters.

Calculus is amazing. Much easier than most methods to solve problems like this.

7 0
3 years ago
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