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marshall27 [118]
3 years ago
10

Liam has purchased a fee-for-service health insurance plan from Leroux Health Insurance. Plan A includes a $248. 00 monthly prem

ium and an annual deductible of $5,500. 00 (not including co-pays). Liam is a fairly healthy young man. He visits his primary care physician three times a year on scheduled visits. He sees a chiropractic specialist every week to help with lower back pain. He has no current prescriptions that need refilling regularly. Assuming Liam maintains his regular visits with his primary care physician and his chiropractor, and avoids any trips to the emergency room or urgent care, how much will Liam pay in health care related fees this year? Leroux Health Insurance Plan A Cost: Monthly Premium: $248. 00 Annual Deductible: $5,500. 00 Co-pays: Brand-name Prescriptions $35. 00 Generic Prescriptions $15. 00 Visits: Primary Care Physician: $40. 00 Specialist: $60. 00 Urgent Care: $125. 00 Emergency Room: $325. 00 a. $716. 00 b. $3,816. 00 c. $5,500. 00 d. $6,216. 0.
Business
1 answer:
Firlakuza [10]3 years ago
6 0

Liam would have to pay $6216 in health care related fees this year. Option D is right

The total expenses that Liam's health would incur are:

Monthly premium at 248 * 12 months in a year

= 2976

52 visits to the specialist at 60 dollars

= 52*60

= $3120

3 Visits to the primary care physician at 40 dollars

= 3*40

= $120

Then Liam's total expenses would be =

$2976 + $120 + $3120

= $6216

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A metal sphere of radius 15 cm has a net charge of 3.0 $ 10#8 C. (a) What is the electric field at the sphere’s surface? (b) If
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Answer:

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Explanation:

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solution

we get here electric field at the sphere’s surface that is

electric field at the sphere’s surface E  =  \frac{q}{4\pi \epsilon _o R^2}   ............1

put here value

electric field at the sphere’s surface E  = \frac{3\times 10^{-8}\times 8.99\times 10^9}{ 0.15^2}    

E = 1.20 × 10^{4} N/C

and

potential on surface of sphere is

V =  \frac{q}{4\pi \epsilon _o R} ................2

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V = 1800 V

and

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ΔV = V(x) - V   ..............3

substitute here value

-500V = \frac{q}{4\pi \epsilon _o } \times (\frac{1}{R+x} - \frac{1}{R})

-500 V = {3\times 10^{-8}\times 8.99\times 10^9} \times (\frac{1}{0.15+x} - \frac{1}{0.15})

solve it we get x

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3 0
3 years ago
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