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3 years ago

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People in a certain group have a 0.60% chance of dying this year. If a person in this group buys a life insurance policy for $

5500 that pays $1,000,000 to her family if she dies this year and $0 otherwise, what is the expected value of the policy?

1 answer:

Nonamiya [84]3 years ago

5 0

Answer:

Explanation:

The expected value is calculated by using the probability of each event. If the chance of dying is 0.60% then the chance living is 99.40%. The expect value formula is:

∑[(xi)*P(xi)] (for all i events).

In this problem we have two events: live or die. If the person dies the family receives $1,000,000 (X1=$1,000,000) and if the person lives the family receives $0 (X2=$0). The probability of receiving $1,000,000 is 60% (P(x1)=0.006) and the probability of receiving $0 is 99.40% (P(x2)=0.994)

Using the formula the expected value of the policy (without the insurance cost):

$1,000,000* (0.006)+ $0*(0,994)= $6,000

If we subtract the insurance value:

$6,000-$5,500= $500

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Pressure with Two Liquids, Hg and Water. An open test tube at 293 K is filled at the bottom with 12.1 cm of Hg, and 5.6 cm of wa

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Answer:

$1170839.28 dyn/cm^2$

16.9816 psia

$117.083928 kN/m^2$

Explanation:

To calculate the absolute pressure in the bottom of tube we need to sum the atmosferic and gauge pressure.

$P_{abs}=P_{atm}+P_G$

And the gauge pressure is given by the contributions of columns of water ( $P_{w}$ ) and mercury( $P_{Hg}$ ), we can calculate the contribution of each column as:

$P= \rho g h$ (*)

where $\rho$ is the respective density, g gravity and h is height.

So we have all the data required to use the above equations ( $P_{atm}$ , height and density of each column) we only need to be carefully with the units.

For simplicity we can to express all pressure contributions in mmHg ( $P_{atm}$ , $P_{w}$ and $P_{Hg}$ ). Note that the units "x" mmHg means the pressure at the bottom of a column of mercury of "x" mm high. For example, in this case we have a column 12.1 cm of Hg, that is a column of 121 mmHg (passing from cm to mm only requires multiply by 10) pressure exerted by that column is 121 mmHg.

Now pressure of 5.6 cm (56 mm) of water would be 56 mm of water, but it is not the same that mmHg, since the density of water is lower, the pressure exerted by 1 mm of water is lower than the exerted by 1 mm of Hg. The conversion between mmHg and mm of water is given by the relation between the densities.

$mmHg=\frac{\rho_w*mmH_2O}{\rho_{Hg}}$

$mmHg=\frac{0.998*mmH_2O}{13.55}=0.0737 mmH_2O$

And pressure of water in mmHg is

$0.0737*56=4.1246 mmHg$

The absolute pressure is:

$P_{abs}=P_{atm}+P_G= 756 + 121 + 4.1246 = 881.1246 mmHg = 88.11246cmHg$

To pass to dyn/cm^2 units we need to use the equation (*)

$P= \rho g h = 13.55 \frac{g}{cm^3} * 980.665 \frac{cm}{s^2} * 88.11246 cmHg = 1170839.28 \frac{g}{cm s^2} = 1170839.28 \frac{dyn}{cm^2}$

Note: We need to use cm Hg for units coherence

Now passing from dyn/ $cm^2$ to kN/ $m^2$ (or kPa) we need to consider that 1 dyn is $10^{-8}$ kN and 1 $cm^2$ is $10^{-4} m^2$ .

$1170839.28 \frac{dyn}{cm^2} * \frac{10^{-8}kN}{1 dyn}*\frac{cm^2}{10^{-4}m^2}=117.083928kN/m^2$

Now passing kN/ $m^2$ to psia. We need to consider that 1 psia is 6.89476.

$117.083928kN/m^2*\frac{1psia}{6.89476kN/m^2}=16.9816 psia$

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Which of the following is not a type of savings account? A. Certificate of savings B. Basic savings C. Savings bond D. Money mar

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A. certificate of savings

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Use the following selected information from Wheeler, LLC to determine the 2017 and 2016 trend percentages for net sales using 20

babunello [35]

Answer:

119.4% for 2017 and 100.0% for 2016.

Explanation:

2017 2016

Net sales $276,200 $231,400

Cost of goods sold $151,900 $129,590

Operating expenses $55,240 $53,240

Net earnings $27,820 $19,820

since we are using 2016 as a base year, the $231,400 in net sales represent 100%, so the trend percentage for 2017 = net sales 2017 / net sales 2016 $276,200 / $231,400 = 1.1936 = 119.4% or a 19.4% increase.

The base year's amount will always be 100% or 1, and the trend percentages will change relative to that year.

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2 years ago

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If a company reorganizes its operation to gain efficiency, the cost associated with this reorganization is classified as

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Answer: Restructuring cost

Explanation:

Restructuring cost could be described as making expenses on rejuvenating or reviving or rebranding the company through spendings, which affects most of it's mode of operations, brings a change and innovation and ways to improve existing methods. This is capital intensive due to the work and changes required during the process.

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On January 1, 2018, Allgood Company purchased equipment and signed a six-year mortgagenote for $186,000 at 15%. The note will be

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Answer:

The correct answer is A: interest= $21048

Explanation:

An amortization schedule is a complete table of periodic loan payments, showing the amount of principal and the amount of interest that comprise each payment until the loan is paid off at the end of its term. While each periodic payment is the same amount early in the schedule, the majority of each payment is interest; later in the schedule, the majority of each payment covers the loan's principal.

Each payment is the same ($49,148), but the proportions of interest and capital pay changes. The interest proportion decreases from pay to pay.

Loan= 186000

i= 15%

n= 6 years

First pay:

i=186000*0,15=27900

amortization= 49148-27900=21248

Second pay:

i=(186000-21248)*0,15=24712

amort=49148-24712=24436

Third pay:

i=(164752-24436)*0,15=21048

amort=49148-21048=28100

While payments progress, interest decreases and amortization increases.

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2 years ago