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3 years ago

A home costing 540,000 is worth 588,000 two years later. What is the one year appreciation rate

Mathematics

2 answers:

gulaghasi [49]3 years ago

5 0

Answer:

A home costing $540,000 is worth $588,000 two years later. What is the one-year appreciation rate? 4.4% is the answer

Step-by-step explanation:

Alenkinab [10]3 years ago

3 0

Answer:

the answer is 48,000

Step-by-step explanation:

its simple you just subtract the bigger number to the smaller number

588,000- 540,000

=48,000 (sorry if it wrong i am not exactly sure sorry please forgive me)

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-3 1/6 + 2 7/9 + 5 2/3 estimated

adelina 88 [10]

Answer:

5.28 if you round to the nearest hundredth. Drop a best answer and a like if it helped

3 0

3 years ago

A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of

notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.01 = 0.99$ , so Z = 2.327.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031$

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.0001 = 2.327\frac{0.0003}{\sqrt{n}}$

$0.0001\sqrt{n} = 2.327*0.0003$

$\sqrt{n} = \frac{2.327*0.0003}{0.0001}$

$(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2$

$n = 48.73$

Rounding up

49 measurements are needed.

7 0

3 years ago

Find the product or quotient. Express using positive exponents

Alex73 [517]

Answer:

Step-by-step explanation:

$x^{-6}*x^{-5}=x^{(-6)+(-5)}=x^{-11}=\frac{1}{x^{11}}$

$x^{a}*x^{b}=x^{a+b}\\\\x^{-a}=\frac{1}{x^{a}}$

6 0

4 years ago

(URGENT!!!) Can someone help me with this? For part a), would the equation A=200(15)^m=395 work? In part b), can't I just factor

Levart [38]

A.) 200+15m=395. This is because the initial price is 200, and after that she will add $15/month. The equation you chose would be exponential, meaning the rate would increase as time passed, which is not the case. Instead, the rate is constant, and the only think that changes is m, the number of months.

b.) I believe it will be easier to answer with ^this equation.

c.) The rate of change would be 15, since the rate is increasing monthly by $15.

d.) Once you solve 209+15m=495, you'll have the answer.

7 0

3 years ago

The equation a=1/2(b^1+b^2)h can be determined the area, a, of a trapezoid with height, h, and base lengths, b^1 and b^2 Which a

Evgesh-ka [11]

The complete question is as follows.

The equation a = $\frac{1}{2}(b_1 + b_2 )h$ can be used to determine the area , a, of a trapezoid with height , h, and base lengths, $b_1$ and $b_2$ . Which are equivalent equations?