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soldier1979 [14.2K]
3 years ago
8

A structure's reproduction cost new is $489,900. The structure has an effective age of 4 years and was built 10 years ago. Assum

ing an economic life of 60 years, how much is the estimated annual depreciation?
Mathematics
1 answer:
olga_2 [115]3 years ago
5 0

Answer:

Annual depreciation will be $32660

Step-by-step explanation:

We have given effective year = 4 years

Economic life = 60 years

Reproduction cost of the structure = $489900

We have to calculate the annual depreciation

Annual depreciation is given by

Annual depreciation =\frac{effective\ age}{economic\ life}\times reproduction\ cost=\frac{4}{60}\times 489900=$32660

So annual depreciation will be $32660

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2+5(8-5) what is solution
gtnhenbr [62]

Answer:

Step-by-step explanation:

brackets: 8-5 = 3

Multiply: 5 * 3 = 15

Add: 2 + 17 = 19

19

6 0
3 years ago
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The distribution of the number of siblings for students at a large high school is skewed to the right with mean 1.8 siblings and
morpeh [17]

Answer:

E. Approximately normal with standard deviation less than 0.7 sibling

Step-by-step explanation:

To solve this question, we use the Central Limit theorem.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

Skewed right distribution, with \mu = 1.8, \sigma = 0.7

Sampling distribution of the sample mean for samples of size 100

By the Central Limit Theorem, they will be approximately normal, with mean \mu = 1.8, and standard deviation s = \frac{0.7}{\sqrt{100}} = 0.07

So the correct answer is:

E. Approximately normal with standard deviation less than 0.7 sibling

4 0
3 years ago
Giovanna used the calculations below to determine the height of a stack of 7 books that are each 2 and StartFraction 5 over 8 En
igor_vitrenko [27]

The error Giovanna made in her calculation is multiplying 7 by 2 and adding with 5/8(step 3)

<h3>Fraction</h3>

Her calculation:

  • 7 (2 and StartFraction 5 over 8 EndFraction): 7(2 5/8)

  • 7 (2 + StartFraction 5 over 8 EndFraction): 7(2 + 5/8)

  • 14 + StartFraction 5 over 8 EndFraction: 14 + 5/8

  • 14 and StartFraction 5 over 8 EndFraction: 14 5/8

Her error was multiplying 7 by 2 and adding with 5/8(step 3)

Correct calculation:

7(2 + 5/8)

= 14 + 35/8

= (112+35) /8

= 147/8

= 18 3/8

Learn more about fraction:

brainly.com/question/11562149

#SPJ1

7 0
2 years ago
***Will mark all right answers brainliest*** A certain type of bacteria is being grown on a Petri dish in the school’s biology l
alex41 [277]

Answer:

On day 0 (starting day), the percentage of petri dish occupied by bacteria was 2.44%

Step-by-step explanation:

Rate of growth = 2  (i.e. doubles every day)

Petri dish was filled to 100% on day 12.

Let

P(0) = percentage of Petri dish occupied on day 0, then

equation of percentage a  function of time in x days

P(x) = P(0)*r^x  ......................(1)

where

100% = P(12) = p(0) * 2^12 = 4096 P(0)

=>

P(0) = 100% / 4096 = 0.0244%

Next, to find percentage on February 14 (Valentine's day!)

Day 0 is February 9, so February 14 is the fifth day, so x=5.

Substitute x=5 in equation (1) above,

P(x) = P(0)*r^x  

P(5) = P(0)*2^5

P(5) = 0.0244*2^5 = 0.0244*32 = 0.781%

Ans. the 0.781% of the petri dish was filled with bacteria after 5 days on February 14th.

4 0
4 years ago
Read 2 more answers
Answer to this question
seropon [69]
The answer is A hope this helps
4 0
4 years ago
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