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2 years ago

PLEASE HELP QUICKLY!!!

Mathematics

2 answers:

weqwewe [10]2 years ago

4 0

Answer:

The Regression line is that line , which passes through all the points , through a single point , or none of the points.

Here, the table is about ,weight of a dog in pounds at various ages in months.

The ratio of Weight at different stages,of months should be proportional.

So, if you look at option (1) , the ratio of x to y in three cases are same that is

$\frac{4}{5}=\frac{8}{10}=\frac{12}{15}$

It means Regression line is Passing through three points, means Dog's weight at three different stages of month are same.and in other tables the ratio are not equivalent.

So, Table 1, that is option 1

is true.

x: 3, 4, 6, 8, 9, 12

y: 2, 5, 7, 10, 11, 15

lyudmila [28]2 years ago

3 0

Answer:

A. x: 3, 4, 6, 8, 9, 12 y: 2, 5, 7, 10, 11, 15

Step-by-step explanation:

I took the test.

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Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressur

Ostrovityanka [42]

Answer:

1) K = 7.895 × 10⁻⁶

2) 0.3024

3) 3.6775 × 10⁻²

4) $f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}$

5) X and Y are not independent variables

$h(x\mid y) = \frac{38000x^2+38000y^2}{3y^2+19000}$

7) 0.54967

8) 25.33 psi

σ = 2.875

Step-by-step explanation:

1) Here we have

$f(x, y) =\begin{cases} & \text (x^{2}+y^{2}) \right. 20\leq x\leq 30 & \ 0 \, Otherwise\end{cases}$

$\int_{x}\int_{y} f(x, y)dydx = 1$

$\int_{x}( \right )\int_{y} f(x, y)dy)dx = 1$

$K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx = 1$

$K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})dx = 1$

$K\int_{x}( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})dx = 1$

$K\int_{x}( (10x^{2})+\frac{19000}{3})_{20}^{30})dx = 1$

$K( (10\frac{x^{3}}{3})+\frac{19000}{3}x)_{20}^{30})= 1$

$K( (10\frac{30^{3}-20^{3}}{3})+\frac{19000}{3}(30-20)))_{20}^{30}) = 1$

$K =\frac{3}{380000}$

2) The probability that both tires are underfilled

P(X≤26,Y≤26) =

$\int_{20}^{26} \int_{20}^{26}K(x^{2}+y^{2})dydx$

$=K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx$

$= K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{26})dx$

$K\int_{x}( (x^{2}(26-20)) +\frac{26^{3}-20^{3}}{3})_{20}^{26})dx$

$K\int_{x}( (6x^{2})+\frac{9576}{3})_{20}^{26})dx$

$K( (6\frac{x^{3}}{3})+\frac{9576}{3}x)_{20}^{26})$

$K( (6\frac{26^{3}-20^{3}}{3})+\frac{9576}{3}(26-20)))_{20}^{26})$

$38304\times K =\frac{3\times38304}{380000}$

= 0.3024

That is P(X≤26,Y≤26) = 0.3024

3) The probability that the difference in air pressure between the two tires is at most 2 psi is given by

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, $| x-y |$ ≤ 2}

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, $\sqrt{(x-y)^2}$ ≤ 2}

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, y ≤ x - 2}

Which gives

20 ≤ x ≤ 22 :: 20 ≤ y ≤ x + 2

22 ≤ x ≤ 28 :: x - 2 ≤ y ≤ x + 2

28 ≤ x ≤ 30 :: x - 2 ≤ y ≤ 30

From which we derive probability as

P( $| x-y |$ ≤2) = $\int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx$ + $\int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx$ + $\int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx$

= K ( $\int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx$ + $\int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx$ + $\int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx$ )

= $K\left [ \left (\frac{14804}{15} \right )+\left (\frac{8204}{15} \right ) +\left (\frac{46864}{15} \right )\right ] = \frac{3}{380000}\times \frac{69872}{15} =\frac{4367}{118750}$ = 3.6775 × 10⁻²

4) The marginal pressure distribution in the right tire is

$f_{x}\left ( x \right )=\int_{y} f(x ,y)dy$

$=K( \right )\int_{y}(x^{2} +y^{2})dy)$

$= K( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})$

$K( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})$

$K(10x^{2}+\frac{19000}{3})}$

$\frac{3}{38000} (10x^{2}+\frac{19000}{3})}$

$= \frac{1}{20} +\frac{3x^{2} }{38000}$

$f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}$

5) Here we have

The product of marginal distribution given by

$f_x(x) f_y(y) = ( \frac{1}{20} +\frac{3x^{2} }{38000})( \frac{1}{20} +\frac{3y^{2} }{38000})$ = $\frac{(3x^2+1900)(3y^2+1900)}{1444000000}$

≠ f(x,y)

X and Y are not independent variables since the product of the marginal distribution is not joint probability distribution function.

6) Here we have the conditional probability of Y given X = x and the conditional probability of X given that Y = y is given by

$h(y\mid x) =\frac{f(x,y))}{f_{X}\left (x \right )}$ = Here we have

$h(y\mid x) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3x^2}{38000} } = \frac{38000x^2+38000y^2}{3x^2+19000}$

Similarly, the the conditional probability of X given that Y = y is given by

$h(x\mid y) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3y^2}{38000} } = \frac{38000x^2+38000y^2}{3y^2+19000}$

7) Here we have

When the pressure in the left tire is at least 25 psi gives

$K\int\limits^{25}_{20} \frac{38000x^2+38000y^2}{3x^2+19000} {} \, dx$

Since x = 22 psi, we have

$K\int\limits^{25}_{20} \frac{38000\cdot 25^2+38000y^2}{3\cdot 25^2+19000} {} \, dx = K \int\limits^{25}_{20} 10.066y^2+6291.39, dx = 57041.942\times \frac{3}{380000}$ = 0.45033