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kobusy [5.1K]
3 years ago
13

During a recent housing slump, the median price of homes decreased in the United States. If James bought his house for $360,000

and the value 1 year later was $253,800, compute the percent decrease in the value of the house.
If necessary, round to the nearest tenth of a percent.
Mathematics
1 answer:
Dahasolnce [82]3 years ago
4 0

Purchase price of the house = $360,000

Price of the house after 1 year = $253,800

Percent decrease in the value of the house = \frac{Purchase price of the house - Price of the house after 1 year}{Purchase price of the house} * 100

⇒ Percent decrease in the value of the house = \frac{360,000 - 253,800}{360,000} * 100

⇒ Percent decrease in the value of the house = \frac{106,200}{360,000} * 100

⇒ Percent decrease in the value of the house = 29.5%

Hence, the price of the house decreased by 29.5% over an year.

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  (c)  162 cm

Step-by-step explanation:

The centroid divides a median into parts with the ratio 1:2, so RL:LD = 1:2. Then RL:RD = 1:(1+2), and RD = 3RL.

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2 years ago
Assume that there is a 4​% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive wit
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Answer:

a) 99.84% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

b) 99.999744% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

Step-by-step explanation:

For each disk drive, there are only two possible outcomes. Either it works, or it does not. The disks are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

4​% rate of disk drive failure in a year.

This means that 96% work correctly, p = 0.96

a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk​ drive, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?

This is P(X \ geq 1) when n = 2

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{2,0}.(0.96)^{0}.(0.04)^{2} = 0.0016

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0016 = 0.9984

99.84% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

b. If copies of all your computer data are stored on four independent hard disk​ drives, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?

This is P(X \ geq 1) when n = 4

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{4,0}.(0.96)^{0}.(0.04)^{4} = 0.00000256

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.00000256 = 0.99999744

99.999744% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

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Therefore, 3/5 and 1/2 are not equivalent.

Hoped this helped.

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