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daser333 [38]
2 years ago
8

Eva is 29 years old and has 2 children, ages 3 and 5.  She makes $48,500 a year.  Eva decides to buy a $400,000 10-year term pol

icy and then renew the policy for another ten years afterwards.  To renew the policy the insurance company charges an extra 40% to her premium rate.  Given the options below, assess whether Eva made a wise decision. a. Eva would have been better off selecting the 20-year term policy. b. Even with the extra charge for renewal, Eva’s plan is the least expensive. c. Given that Eva plans to renew, she should have selected the whole life policy. d. Eva ends up paying the same amount for each policy.
Mathematics
2 answers:
nirvana33 [79]2 years ago
8 0
I believe it's A. <span>Eva would have been better off selecting the 20-year term policy.
Unde current circumtances, 10-year term policy wouldn't guarantee thesafety of the kids because even after the policy ends, Eva's kids still haven't entered the age where they could find their own income (they would be 12, 13, and 15).
If Eva decided to add another 10 year despite the extra charge, The kids will be covered until they enter the productive age.</span>
patriot [66]2 years ago
6 0

Answer with explanation:

Description about Eva:

1. She is 29 years old, and has 2 children having ages 3 and 5.

2. She decided to buy , a $400,000 10-year term policy and then renew the policy for another ten years afterwards.  

3. For renewal of policy after 10 years , she has to pay 40 % extra.

The present value of money and value of money after 10 years will change, so giving extra 40 % after 10 years is not an issue.

The financial condition of Eva presently is fine, and we can't predict her future, as her children's are growing so after 10 years what will be her financial condition , she is not sure. So, Decision taken by Eva is surely sensible and sharp witted.

Option B:   Even with the extra charge for renewal, Eva’s plan is the least expensive

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A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si
Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

Area = 128

Let the dimension of the paper be x and y;

Such that:

Length = x

Width = y

So:

Area = x * y

Substitute 128 for Area

128 = x * y

Make x the subject

x = \frac{128}{y}

When 1 inch margin is at top and bottom

The length becomes:

Length = x + 1 + 1

Length = x + 2

When 2 inch margin is at both sides

The width becomes:

Width = y + 2 + 2

Width = y + 4

The New Area (A) is then calculated as:

A = (x + 2) * (y + 4)

Substitute \frac{128}{y} for x

A = (\frac{128}{y} + 2) * (y + 4)

Open Brackets

A = 128 + \frac{512}{y} + 2y + 8

Collect Like Terms

A = \frac{512}{y} + 2y + 8+128

A = \frac{512}{y} + 2y + 136

A= 512y^{-1} + 2y + 136

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

A' = -512y^{-2} + 2

Set

A' = 0

This gives:

0 = -512y^{-2} + 2

Collect Like Terms

512y^{-2} = 2

Multiply through by y^2

y^2 * 512y^{-2} = 2 * y^2

512 = 2y^2

Divide through by 2

256=y^2

Take square roots of both sides

\sqrt{256=y^2

16=y

y = 16

Recall that:

x = \frac{128}{y}

x = \frac{128}{16}

x = 8

Recall that the new dimensions are:

Length = x + 2

Width = y + 4

So:

Length = 8 + 2

Length = 10

Width = 16 + 4

Width = 20

To double-check;

Differentiate A'

A' = -512y^{-2} + 2

A" = -2 * -512y^{-3}

A" = 1024y^{-3}

A" = \frac{1024}{y^3}

The above value is:

A" = \frac{1024}{y^3} > 0

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

3 0
1 year ago
in a basketball game tatiana made 23 baskets . each of the baskets was worth either 2 or 3 points and Tatiana scored a total of
natima [27]
Here is the answer to you question!

8 0
1 year ago
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A survey of 132 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take
ddd [48]

Answer:

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 66/132 = 0.50

Number of samples n = 132

Confidence level = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

0.50 +/- 1.645√(0.50(1-0.50)/132)

0.50 +/- 1.645√(0.001893939393)

0.50 +/- 0.071589436011

0.50 +/- 0.072

(0.428, 0.572)

The 90% confidence level estimate of the true population proportion of students who responded "yes" is (0.428, 0.572)

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

7 0
2 years ago
Ann took a taxi home from the airport. The taxi fare was \$2.10$2.10dollar sign, 2, point, 10 per mile, and she gave the driver
Studentka2010 [4]

Answer:

2.10x+5=49.10

Distance = 21 miles.  

Step-by-step explanation:

Let x be the distance in miles between the airport and Ann's home.

We have been given that Ann took a taxi home from the airport. The taxi fare was $2.10 per mile. So fare for x miles will be 2.10x.

We are also told that she gave the driver a tip of $5. Ann paid a total of $49.10. This means that fare of x miles and tip given to driver equals $49.10. We can represent this information in an equation as:

2.10x+5=49.10

Now let us solve our equation.

2.10x=49.10-5

2.10x=44.10

x=\frac{44.10}{2.10}=21

Therefore, the equation 2.10x+5=49.10 represents the distance in miles between the airport and Ann's home and distance between Ann's home and airport in 21 miles.


3 0
2 years ago
Read 2 more answers
An animal park has lions, tigers and zebras. 20% of the animals are lions and half of the animals are zebras. If there are 120 a
RoseWind [281]

Answer:

36\ tigers

Step-by-step explanation:

Let

x ----> the number of lions

y ----> the number f tigers

z ----> the number of zebras

we know that

x+y+z=120 ----> equation A

Remember that

20\%=20/100=0.20

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To find out the number of lions, multiply the total animals by the percentage of lions

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To find out the number of zebras, multiply the total animals by the percentage of zebras

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Substitute the value of x and the value of z in equation A and solve for y

24+y+60=120

y+84=120

y=36\ tigers

6 0
2 years ago
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