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ozzi
2 years ago
8

. A condo in Orange Beach, Alabama, listed for $1.4 million with 20% down and financing at 5% for 30 years. What is the monthly

payment?
Mathematics
1 answer:
vovikov84 [41]2 years ago
8 0
Listed price = $1.4 million
Down payment = 20% of $1.4 million = 0.2 x 1,400,000 = 280,000
Amount left to pay = $1.4 million - 280,000 = $1,120,000
Present value of an annuity is given by PV = P(1 - (1 + r/t)^-nt) / r
where: PV = $1,120,000
r = 5% = 0.05
t = 12
n = 30 years.
1,120,000 = P(1 - (1 + 0.05/12)^-(12 x 30)) / 0.05
1,120,000 x 0.05 = P(1 - (1 + 1/240)^-360)
56,000 = P(1 - 0.2238)
P = 56,000 / 0.7761 = 72,148.83
Therefore, the monthly payment is $72,148.83
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(a) Yes, because the probabilities add up to 1.

(b) The probability that <em>X</em> < 40 is 0.80.

Step-by-step explanation:

The probability distribution of the random variable <em>X</em> is:

<em>    x</em>:  15   |   22  |   34   |   40

f (x): 0.14 | 0.40 | 0.26 | 0.20

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The properties of a probability distribution are:

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All the probability value are more than 0 and less than 1.

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The sum of all probabilities is 1.

Thus, the probability distribution is valid.

(b)

Consider the probability distribution table.

Compute the probability of <em>X</em> < 30 as follows:

P (X < 40) = P (X = 15) + P (X = 22) + P (X = 34)

                =0.14+0.40+0.26\\=0.80

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1) The linear regression model is y = -0.0348·x + 13.989

2) The correlation coefficient is -0.0725

3) The strength of the model is strong - association

Step-by-step explanation:

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                         X            Y          XY       X²

                         27            13           351          729

                         65             12          780         4225

                         83              11          913         6889

                         109            10          1090      11881

                         142            9            1278     20164

                         175              8           1400      30625

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a = \frac{n\sum xy - \sum x\sum y }{n\sum x^{2}-\left (\sum x  \right )^{2}} = \frac{6 \times 5812  - 601 \times 63}{6 \times 74513-601^{2}} = - 0.0348

b = 1/n(∑y -a∑x) = 1/6(63 - (0.0348)×601) = 13.989

Therefore, the linear regression model is y = -0.0348·x + 13.989

2)

r = \frac{n\sum xy - \sum x\sum y }{\sqrt{[n\sum x^{2}-\left (\sum x  \right )^{2}] [n\sum y^{2}-\left (\sum y  \right )^{2}]}}  = \frac{6 \times 5812  - 601 \times 63}{\sqrt{[6 \times 74513-601^{2}] [6  \times 3969 - 63^2]} } = - 0.0725

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