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elena55 [62]
3 years ago
15

Kevin, who has liability limits of $20,000/$40,000/$20,000 and a $500 collision deductible, is involved in an accident. Due to h

is negligence he runs into a Rolls Royce. Three people sustained bodily injuries in the Rolls Royce. The driver's injuries were worth $1,000,000, a passenger received injuries worth $12,500, and another passenger received injuries of $7,100. How much will the PAP pay for these bodily injuries?
Business
1 answer:
harina [27]3 years ago
4 0

Answer:

Amount to pay by PAP = $39,600

Explanation:

The liability limits of $20,000/$40,000/$20,000 implies that the highest amount PAP will pay for driver's injuries is $20,000, while the highest to pay for the first of two passenger is $40,000 and $20,000 for second passenger.

Since the a passenger received injuries worth $12,500, and another passenger received injuries of $7,100, the PAP will the actual amount and $20,000 for the driver's injuries. The total can therefore be calculated as follows:

Amount to pay by PAP = $20,000 + $12,500 + $7,100 = $39,600

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Production Oriented Era can be regarded as an era in which manufacturers were concerned with product innovation, they do this instead of meeting customers needs.

In this era Retailers were considered places to hold inventory until it was sold.

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3 years ago
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3 years ago
Which of the following (all other factors held constant) will cause an increase in a stock’s value?
Lady_Fox [76]

Answer:

B, a decrease in the stock's beta.

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Mostly, stocks with a volatility below 1.0 is less volatile compared to stocks with a beta above 1.0.

The beta of a stock is calculated by finding the rate, the rate of return and the market rate of return of the stock. All of these above are to expressed as a percentage. Having gotten the percentages from above, the risk free rate is subtracted from the rate of return of the stock. After that, the risk free rate is also subtracted from the market rate of return.

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Cheers.

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3 years ago
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Yuliya22 [10]

Answer:

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Data provided as per the question below:

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