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quester [9]
2 years ago
14

Tom Worker pays $900 annually for $60,000 worth of life insurance

Mathematics
1 answer:
Gelneren [198K]2 years ago
4 0

Answer:

a) = $9000  10yr premiums

b) = $6016.95  Cash value

c) Ratio = 2.017 : 3   =  67%  Ratio Value To Paid Premiums in percent.

Step-by-step explanation:

The life insurance has a 5 year changeable rate after every 5 years.

So we need to find ratio every 5 years to work out what he was paid.

Instead of working out each one

We look for patterns for 60,000

If 5 year = 60,000 this means 2x

But it was 10 years 89/30 =2.9666...rep x (counter-check multiplier)

Back to questions

Tom paid 900 x 10 = 9,000 = ? in value

Answer a = What did he pay = $9000

Answer b = What is this $9000 in cash value where 5 years shows 30,000 the ratio is 9:30 = 3:10 so for 5 years this would = 30% of 30,000 =$9,000 for 5 years.

For 10 years the ratio goes up from 30% of 30,000 to 30,000 of 89,000 = 33.71

To prove 9000 is cash share of 30,000 at 5 years we have to half = $4500 for 5 years.

To prove 9000 is cash share of 89,000 at 10 years we take the $4500 and multiply it by 33.71% = 4500 x 1.3371 = 6016.95

b)  So his cash value = $6016.95

His life insurance = $60,000 as told

c) The ratio = 6016.95 : 9000 simplify = 2.017 : 3  (as we rounded and added + $16.95 as 0.017. for cash value to paid premiums

We  (counter-check multiplier) 2.9666

9000/2.9666 = 3033.77603991

9000- 6016.95 = 2983.05

So we can see after 5 years it decreased by $40

But for b and c, lets say we didnt half it 9000 = 30,000 = 30% 33.71 = 33.71%

We then see 30% of 89000 = 30,000 and x 9000 by 1.33.71 = $12033.90

So cash value will show 10 years = $12,033.90

Showing ratio 9000: 12033.90 =

9000/33.90 =265 = 2.65

So 75% - 2.65% = 3% rounded = 72%

This inverse = 28%

Then the percentage = 67.% can be changed to 28% or used as 72%

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Answer:

0.75 = 1-\frac{1}{k^2}

If we solve for k we can do this:

\frac{1}{k^2}= 1-0.75=0.25

\frac{1}{0.25}= k^2

k^2 =4

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So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

Lower = \mu -2\sigma = 514- 2*40=434

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Step-by-step explanation:

We don't know the distribution for the scores. But we know the following properties:

\mu = 514 , \sigma =40

For this case we can use the Chebysev theorem who states that "At least 1 -\frac{1}{k^2} of the values lies between \mu -k\sigma and \mu +k\sigma"

And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:

0.75 = 1-\frac{1}{k^2}

If we solve for k we can do this:

\frac{1}{k^2}= 1-0.75=0.25

\frac{1}{0.25}= k^2

k^2 =4

k =\pm 2

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

Lower = \mu -2\sigma = 514- 2*40=434

Upper = \mu +2\sigma = 514 + 2*40=594

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The number of messages that arrive at a Web site is a Poisson distributed random variable with a mean of 6 messages per hour. Ro
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Answer:

a) There is a 16.0623% probability that 5 messages are received in 1 hour.

b) There is a 11.5880% probability that 10 messages are received in 1.5 hours.

c) There is a 22.4042% probability that 2 messages are received in 0.5 hours.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

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So

P(X = 5) = \frac{e^{-6}*(6)^{5}}{(5)!} = 0.160623

There is a 16.0623% probability that 5 messages are received in 1 hour.

(b) What is the probability that 10 messages are received in 1.5 hours?

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There is a 11.5880% probability that 10 messages are received in 1.5 hours.

(c) What is the probability that less than 2 messages are received in 1/2 hour?

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