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2 years ago

Which expressions is equivalent -8.2-5

Mathematics

1 answer:

GuDViN [60]2 years ago

8 0

Answer:

-13.2

Step-by-step explanation:

-8.2-5 essentially means that you just have to add 8.2 and 5. (disregard the signs for a moment) if you add them you get 13.2 as both of these number have a negative sign just add it and here's your answer

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What is the solution to the equation 1 over 56 multiplied by x is equal to 1 over 8

Kruka [31]

Lets write the equation
x*1/56= 1/8

Lets do left side

1x/56=1/8

Now we see that in order to remove 56 and only be left with 1x on leftside we have to multipley by 56. If we do so on on side we have to do on the other side. So both sides multiplies by 56

1x/56*56= 1*56/8

From above you see that 1x= 7

x=7

5 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