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docker41 [41]
3 years ago
8

This chart shows the total average monthly cost of medical insurance benefits for employees working in the private sector in mar

ch 2006. for single coverage, the cost is $322.77, and for family coverage, it is $889.26. if an employee makes $10 per hour (after withholdings) and has to pay the full cost of single coverage medical insurance, about how many hours will he or she have to work each month simply to pay for the cost of the medical insurance
Mathematics
1 answer:
marusya05 [52]3 years ago
4 0
It is given in this item that the subject of interest has to pay the single coverage for the medical insurance and is given to be equal to $322.77. 

Let x be the number of hours that a person should work in order to pay for the cost of the medical insurance. The value of this variable is calculated by dividing the total insurance cost by the amount that a person earns every hour.

                             x = $322.77 / ($10/hr)
                             x = 32.277 hours

Therefore, the number of hours that a person should work is equal to 32.277 which is approximately equal to 32 hours and 17 minutes. 
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On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or
almond37 [142]

Answer:

a) \vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j, b) \theta = \frac{\pi}{4}, c) y_{max} = 84.375\,m, t = 3.75\,s.

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j

The particular expression for the cannonball is:

\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j

b) The components of the position of the cannonball before hitting the ground is:

x = (90\cdot \cos \theta)\cdot t

0 = 90\cdot \sin \theta - 6\cdot t

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta}  \right)

0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x

0 = 4050\cdot \sin 2\theta - 6\cdot x

6\cdot x = 4050\cdot \sin 2\theta

x = 675\cdot \sin 2\theta

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

\frac{dx}{d\theta} = 1350\cdot \cos 2\theta

1350\cdot \cos 2\theta = 0

\cos 2\theta = 0

2\theta = \frac{\pi}{2}

\theta = \frac{\pi}{4}

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta

\frac{d^{2}x}{d\theta^{2}} = -2700

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

y = 45\cdot t - 6\cdot t^{2}

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

\frac{dy}{dt} = 45 - 12\cdot t

45-12\cdot t = 0

t = \frac{45}{12}

t = 3.75\,s

Now, the second derivative is used to check if such solution leads to a maximum:

\frac{d^{2}y}{dt^{2}} = -12

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}

y_{max} = 84.375\,m

7 0
3 years ago
How many distinguishable 11 letter​ "words" can be formed using the letters in MISSISSIPPI​?
SCORPION-xisa [38]
This is given by the multinomial coefficient:

\dbinom{11}{1,4,4,2}=\dfrac{11!}{1!4!4!2!}=34650

If you're not familiar with the multinomial coefficient, you may be able to see it more clearly if you count the number of possible combinations taking each distinct letter n times, where n is the number of times it shows up in the original word.

\underbrace{\dbinom{11}1}_{\text{M}}\underbrace{\dbinom{10}4}_{\text{I}}\underbrace{\dbinom64}_{\text{S}}\underbrace{\dbinom22}_{\text{P}}=\dfrac{11!}{1!10!}\dfrac{10!}{4!6!}\dfrac{6!}{4!2!}\dfrac{2!}{2!0!}=\dfrac{11!}{1!4!4!2!}
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3 years ago
Amira is solving the equation x2 – 6x = 1. Which value must be added to both sides of the equation to make the left side a perfe
Lorico [155]

Answer:

its 9

Step-by-step explanation:

I have  gotten it correct on ED

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3 years ago
TRUE OR FALSE MATH PLZ HELP:
Sveta_85 [38]

Answer:

question 2 is true

question 3 is false

Step-by-step explanation:

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Find the sum of the first 48 terms of the following series, to the nearest integer. 11, 18,25,...
Zigmanuir [339]

Answer:

8424

Step-by-step explanation:

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