1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
IgorC [24]
3 years ago
14

Bob and Roberta are married and have two children. They are covered by a comprehensive medical and major medical policy with a $

200 deductible per person, an 80/20 coinsurance provision, and an annual stop loss of $3,000. Bob dislocated his shoulder in November and needed surgery to fix the injury, which cost $35,000. Bob, Roberta, and each of the children had an annual physical exam earlier in January, which cost $250 each. Aside from these expenses, there were no other health care costs incurred by the family this year. How much will Bob have to pay out of pocket to cover the costs of his surgery?
Mathematics
1 answer:
Novay_Z [31]3 years ago
4 0

Answer:

Bob will have to pay $2,160 to cover the cost of his surgery

Step-by-step explanation:

Coinsurance on the surgery = 20% * $35,000

Coinsurance on the Surgery = 20/100 * $35,000

Coinsurance on the Surgery = 0.2 * $35,000

Coinsurance on the Surgery = $7,000

Annual Stop Loss = $3,000

Annual Physical Exam costs = $250 * 4

Annual physical Exam = $1000

Coinsurance on physical exam = 80% * $1,000

Coinsurance= 80/100 * $1,000

Coinsurance = 0.8 * $1,000

Coinsurance = $800

Deductible paid for Physical Exam = 20% *$200

Deductible = 20/100 * $200

Deductible = 0.2 * $200

Deductible = $40

Total payables = Annual Loss - Coinsurance on Physical Exam - Deductible for Physical Exam

Total Payables = $3,000 - $800 - $40

Total Payables = $2,160

The insurance company will pay the rest of the medical expenses.

You might be interested in
Write 800 as the product of its prime factors.
Alex_Xolod [135]
800|2
800|2
400|2
200|2
100|2
50|2
25|5
5|5
1

800=2^5\cdot5^2


8 0
3 years ago
A graph shows the horizontal axis numbered 2 to 8 and the vertical axis numbered 10 to 50. A line increases from 0 to 4 then dec
3241004551 [841]

Answer:

very easy A graph shows the horizontal axis numbered 2 to 8 and the vertical axis numbered 10 to 50. A line increases from 0 to 4 then decreases from 4 to 9.

Which type of function best models the data shown on the scatterplot?

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
I’ll be very grateful for anyone’s help I added a photo of the question
Irina18 [472]

Answer:

c

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The student council at Spend too much High School is planning a school dance. The table below shows the budget for the dance.
Pie
The dance would /cost/ the school 50 dollars, making the profit -50. The dance tickets make 300, and the DJ takes away 250, making a 50 profit. However, the decorations cost 100, taking away that from 50, making -50.
4 0
2 years ago
Arrange the following numbers in order from greatest to least 7,361; 7,136; 7,613
d1i1m1o1n [39]
7,613 ,7,361,7,136

 hope that helped
3 0
3 years ago
Read 2 more answers
Other questions:
  • M^2-6m+9\m^2-2m-3÷m^2-5m+6\m^2-3m+2
    12·1 answer
  • Add. write your answer as a fraction in simplistic form<br><br> -4 5/9 + 8/9
    15·1 answer
  • "Can anyone help me?<br><br><img src="https://tex.z-dn.net/?f=helppp%20%5C%3A%20me" id="TexFormula1" title="helppp \: me" alt="h
    10·1 answer
  • A-9=6<br> p-20=-30<br> z-(-12)=15<br> x-(-7)=10
    13·1 answer
  • Two more than the product of a number and three
    5·1 answer
  • What's the temperature on each thermometer?
    6·1 answer
  • Please answer this for me really struggling thanks
    5·1 answer
  • In a bag of marbles, there are 4 blue marbles, 6 yellow marbles, 3 red marbles and 6 green marbles. If you choose a yellow marbl
    10·1 answer
  • Help me plss im beeging
    11·1 answer
  • Please please please help me asap!
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!