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2 years ago

Simplify the expression. (x6y-6) ———— (x8y-8)

1 answer:

3 0

The scatterplot shows how many miles Marissa jogged over 10 days. Which statements are true? Check all that apply.

Marissa jogged 5 miles on the first day.
Marissa jogged 6 miles on the second day.
Marissa jogged 9 miles on the third day.
Marissa jogged 3 miles on the fourth day.
Marissa jogged 3 miles on the ninth day.vvvvvThe scatterplot shows how many miles Marissa jogged over 10 days. Which statements are true? Check all that apply.

Marissa jogged 5 miles on the first day.
Marissa jogged 6 miles on the second day.
Marissa jogged 9 miles on the third day.
Marissa jogged 3 miles on the fourth day.
Marissa jogged 3 miles on the ninth day.Answer:

Step-by-step explanation:
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Which function has an inverse that is also a function?

dimaraw [331]

Answer:

g(x) = 2x – 3 is the function has an inverse that is also a function.

The correct option is A.

Inverse function;

The inverse of a function is the equation where x and y have switched places.

Here g(x) = 2x - 3 has inverse x = 2y - 3 which simplifies to;

$y = \frac{x-3}{2}$

$y = \frac{1}{2} x - \frac{3}{2}$

k(x) = - $9x^{2}$ has the inverse x = $-9y^{2}$ which simplifies to;

$y =\sqrt{ \frac{x}{-9} }$

f(x) = |x+2| is an absolute value function.

w(x) = -20 does not have an inverse which is a function.

Therefore

The function has an inverse that is also a function is g(x) = 2x – 3.

#SPJ1

3 0

2 years ago

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