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?

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.

395.84

step-by-step explanation:

2πrh+2πr²

2 (pie)(radius)(height)+2(pie)(radius)^2

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}