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2 years ago

6 Using data from 1988 for houses sold in Andover, Massachusetts, from Kiel and McClain (1995),

1 answer:

5 0

The study of explanatory or independent differences based on the dependent variable is a simple statistical method. The correlation between two variables is researched in which only the explanatory variable of them would be present, and the further discussion can be defined as follows:

For point (i):

The log(distances) seem to be optimistic, and their valuation is 0.312, which also means that the cost of a house will rise even as the distance increases.
In this, the certainly is what everybody expects the better the house from the sewage treatment is from a hygienic viewpoint.

For point (ii):

In this regard, we cannot use straightforward regression, even as non-biased price estimation might not even be provided.
The price-distance relationship here is not so much linear.

For point (iii):

Several factors influence the house's price, the biggest contributor is house length, market nearness or public areas, number of rooms, etc.
Thus the only influence is not the mere distance from the incinerator. It's one of the factors affecting housing pricing.
The proximity of markets and places can be connected to the distance from the crematorium.
However, factors such as house size and the number of stories aren't related to the distance from the incineration plant.

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A fruit salad recipe requires 4 cups of grapes per 8 servings. How many cups of grapes are required for 28 servings? A) 6 Cups B

Dafna1 [17]

Answer:

D) 14 Cups

Step-by-step explanation:

4 0

2 years ago

1. Three friends pooled their money to purchase a new game system that costs $298. One person

iragen [17]

Answer:

First person: $107

Second person: $98

Third person: $93

Step-by-step explanation:

Let be "f" the amount of money (in dollars) that the first person contributed to the purchase, "s" the amount of money (in dollars) that the second person contributed to the purchase and "t" the amount of money (in dollars) that the third person contributed to the purchase.

With the information given in the exercise, you can set up the following equations:

Equation 1 → $f+s+t=298$

Equation 2 → $s=f-9$

Equation 3 → $t=f-14$

Substitute the Equations 2 and 3 into the Equation 1 and then solve for "f":

$f+(f-9)+(f-14)=298\\\\3f-23=298\\\\f=\frac{321}{3}\\\\f=107$

Finally, substitute the value of "f" into the Equation 2 and then into the Equation 3, in order to find the values of "s" and "t".

Therefore, you get:

$s=107-9\\\\s=98\\\\\\t=107-14\\\\t=93$

5 0

2 years ago

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

2 years ago

Use the rule to write the numbers in the pattern rule:add3the

almond37 [142]

Answer:

Hey! Is this the full question?

Step-by-step explanation:

5 0

2 years ago

Solve the triangle. A = 46° a = 33 b = 26

frozen [14]

Using sin rule:
a/sin A =b/sin B=c/sin c
SIN A=sin 46=0.72
THEN:sin B =(0.72*26)/33=0.57
then B= 34°
AS the sum of triangle angles=180
then C=180-(46+34)=100°
then c =(a*sin C)/sin A
sin C=sin 100=0.98
=(33*0.98)*0.72=45

7 0

3 years ago

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