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boyakko [2]
1 year ago
11

Mary Hernandez had a policy with a $250 deductible which paid 80% of her covered charges less deductible. She had medical expens

es of $18,240.00.
Calculate:

a. the insurance company's payment

b. the 20% copayment

c. Mary's total cost
Mathematics
2 answers:
Nataly_w [17]1 year ago
6 0
A. the insurance company's payment = $14,392
$18240 - $250 = 17990 x 80% = $14,392

b. <span>the 20% copayment
17990 * 20% = $3598
</span>
<span>c. Mary's total cost
$250 + $3598 = $3848
</span>
Ugo [173]1 year ago
6 0

Answer

Given

Mary Hernandez had a policy with a $250 deductible which paid 80% of her covered charges less deductible.

She had medical expenses of $18,240.00.

First find out the insurance company's payment .

Definition of deductible

The insurance deductible is the amount of money you will pay in an insurance claim before the company starts paying you.

Thus

Medical expenses after reducing the deductible =  $18,240 -  $250

                                                                                 = $ 17990

As given

The policy paid  80% of Mary covered charges less deductible.

80% is written in the decimal form.

= \frac{80}{100}

= 0.80

insurance company's payment = 0.80 × 17990

                                                   = $14392

Second find the 20% copayment  

20 % is written in the decimal form.

= \frac{20}{100}

= 0.20

Thus

The 20% copayment  = 0.20 × 17990

                                   = $3598

Third find the Mary's total cost

Mary total cost = Deductible cost +  20% copayment cost

                        = 250 + 3598

                         = $ 3848

Thus the Mary total cost is  $ 3848 .


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There are 200 students in eleventh grade high school class. There are 40 students in the soccer team and 50 students in the bask
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Step-by-step explanation:

P(A) is the probability that the selected student plays soccer.

Then:

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P(B) is the probability that the selected student plays basketball.

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P(B)=\dfrac{50}{200}=0.25

P(A and B) is the probability that the selected student plays soccer and basketball:

P(A\&B)=\dfrac{10}{200}=0.05

P(A|B) is the probability that the student plays soccer given that he plays basketball. In this case, as it is given that he plays basketball only 10 out of 50 plays soccer:

P(A|B)=\dfrac{P(A\&B)}{P(B)}=\dfrac{10}{50}=0.2

P(A | B) is equal to P(A), because the proportion of students that play soccer is equal between the total group of students and within the group that plays basketball. We could assume that the probability of a student playing soccer is independent of the event that he plays basketball.

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The formula for determining the standard error of the distribution of difference in means is expressed as

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5 0
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An open-top rectangular box is being constructed to hold a volume of 150 in3. The base of the box is made from a material costin
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Answer:

minimum cost of construction box are x = x = 3.4199 in and y = 10.2598 in and z = 4.2750 in

Step-by-step explanation:

given data

volume = 150 in³

base material costing = 5 cents/in²

front cost = 10 cents/in²

remainder sides cost = 2 cents/in²

to find out

the dimensions that will minimize the cost

solution

we consider here length = x and breadth = y and height = z

and

area of base = xy

area of front = xz

and area of remaining side = xz + 2yz     .....................1

so

cost of base will be = 5xy

cost of front = 10xz

cost of remaining side = 2 ( xz+ 2yz)        

and

total cost will be

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total cost TC = 5xy + 10xz + 2xz+ 4yz

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5x - \frac{1800}{y^2} = 0

x = \frac{360}{y^2}    

solve we get

y = 10.2598

x = 3.4199

now put x and y value in equation 3

z = \frac{150}{xy}  

z = \frac{150}{3.4199*10.2598}  

z = 4.2750

so minimum cost of construction box are x = x = 3.4199 in and y = 10.2598 in and z = 4.2750 in

4 0
2 years ago
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