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4 years ago

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An insured's roof cost $4,000 when installed 5 years ago. It has been damaged by hail and must be replaced. The new roof will co

st $6,000 at today’s prices. If the roof has been depreciating at $200 per year and the insured’s policy is written on the actual cash value(ACV), how much will the policy pay toward the insured's new roof?

1 answer:

hoa [83]4 years ago

5 0

Answer:

ACV=$4,500

Step-by-step explanation:

We have that the actual cash value (ACV) is defined as:

$ACV=\dfrac{R\times(E-C)}{E}$

Where:

$ACV =$ actual cash value

$R =$ replacement cost or purchase price of the item

$E =$ expected life of the item

$C =$ current life of the item

Then we have R=$6,000, C=5years, and to find the expected life of the item we can use the depreciating of the roof, then if the roof is depreciating $200 each year we just need to divide $4,000 by $200 to find the expected life of the roof:

$\dfrac{4,000}{200}=20$

Then the espected life of the roof is 20 years, with this result we have all the data, then:

$ACV=\dfrac{\$6,000\times (20-5)}{20}=\dfrac{\$6,000\times (15)}{20}=\dfrac{\$90,000}{20}=\$4,500$

Then the ACV is $4,500

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Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below: Team X: 11, 3, 0,

Gre4nikov [31]

Answer:

C. Team Y’s scores have a lower mean value.

Step-by-step explanation:

We are given that Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below:

Team X: 11, 3, 0, 0, 2, 0, 6, 4

Team Y: 4, 2, 0, 3, 2, 1, 6, 4

Firstly, we will calculate the mean, median, range and inter-quartile range for Team X;

Mean of Team X data is given by the following formula;

Mean, $\bar X$ = $\frac{\sum X}{n}$

= $\frac{11+ 3+ 0+ 0+ 2+ 0+ 6+ 4}{8}$ = $\frac{26}{8}$ = 3.25

So, the mean of Team X's scores is 3.25.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team X: 0, 0, 0, 2, 3, 4, 6, 11

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

Here, the number of observations is even, i.e. n = 8.

So, Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}$

= $\frac{2+3}{2}$ = 2.5

So, the median of Team X's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 11 - 0 = 11

So, the range of Team X's score is 11.

Now, the inter-quartile range of the data is given by;

Inter-quartile range = $Q_3-Q_1$

$Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}$

= $(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(2.25 )^{th} \text{ obs.}$

$Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]$

= 0 + 0.25[0 - 0] = 0

$Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}$

= $3(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(6.75 )^{th} \text{ obs.}$

$Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]$

= 4 + 0.75[6 - 4] = 5.5

So, the inter-quartile range of Team X's score is (5.5 - 0) = 5.5.

<u>Now, we will calculate the mean, median, range and inter-quartile range for Team Y;</u>

Mean of Team Y data is given by the following formula;

Mean, $\bar Y$ = $\frac{\sum Y}{n}$

= $\frac{4+ 2+ 0+ 3+ 2+ 1+ 6+ 4}{8}$ = $\frac{22}{8}$ = 2.75

So, the mean of Team Y's scores is 2.75.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team Y: 0, 1, 2, 2, 3, 4, 4, 6

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

Here, the number of observations is even, i.e. n = 8.

So, Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}$

= $\frac{2+3}{2}$ = 2.5

So, the median of Team Y's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 6 - 0 = 6

So, the range of Team Y's score is 6.

Now, the inter-quartile range of the data is given by;

Inter-quartile range = $Q_3-Q_1$

$Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}$

= $(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(2.25 )^{th} \text{ obs.}$

$Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]$

= 1 + 0.25[2 - 1] = 1.25

$Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}$

= $3(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(6.75 )^{th} \text{ obs.}$

$Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]$

= 4 + 0.75[4 - 4] = 4

So, the inter-quartile range of Team Y's score is (4 - 1.25) = 2.75.

Hence, the correct statement is:

C. Team Y’s scores have a lower mean value.

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4 years ago

The soot produced by a garbage incinerator spreads out in a circular pattern. The depth, H(r), in millimeters, of the soot depos

Anon25 [30]

Answer:

Step-by-step explanation:

Check attachment for solution

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3 years ago

The perimeter of triangle ABC is 32 find the lengths of AC and AB

Nutka1998 [239]

The lengths of AC and AB are each 10.6 units.

<u>Step-by-step explanation</u>:

Perimeter of a triangle = Sum of all the three sides of a triangle

Perimeter = 3a
where, a is the length of the each sides of a triangle.

⇒ 32 = 3a

⇒ a = 32/3

⇒ a = 10.6

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4 years ago

Each person in a group of students was identified by year asked

Natalija [7]

Answer:

0.184

Step-by-step explanation:

There are 38 seniors, of which 7 prefer evening classes.

7/38 ≈ 0.184

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3 years ago

Read 2 more answers

This makes no sense to me keep getting the wrong answers when i try to figure it out

satela [25.4K]

To solve this problem, you'd want to start by finding the mean of the given numbers. To find the mean, add all the numbers together and divide by how many there are.

Next, you'll see that the question says one of the rents changes from $1130 to $930. So find the mean of all the numbers again, except include $930 in your calculation instead of $1130.

I got $990 as the mean for the given numbers, and $970 as the mean after replacing the $1130 with $930. Subtracting the two means gives you $20. So the mean decreased by $20.

Now for the median, all you need to do is find the median of the given numbers and compare them with the median of the new data. Because there are ten terms, you have to add the middle two numbers and divide by two. $990 + $1020 = 2010. 2010÷2 = $1005 as the first median.

The new rent is 930, so you have to reorder the data so it goes from least to greatest again. 745, 915, 925, 930, 965, 990, 1020, 1040, 1050, 1120. After finding the median again you get 977.5. Subtracting the two medians gives you $27.5 as how much the median decreased. Hope this helps!

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3 years ago