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Sergio039 [100]

3 years ago

9

1. Each of the integers from 0 to 9, inclusive, is written on a separate slip of blank paper and the ten slips are dropped into

a hat. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?

1 answer:

Ludmilka [50]3 years ago

5 0

Answer:

5, assuming you get the worst possible slips off: 0,1,2,3,4

Step-by-step explanation:

You might be interested in

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

6)

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

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

For P(Y≥25) we have

$K \int\limits^{30}_{25} 10.066y^2+6291.39, dx = 69624.72\times \frac{3}{380000}$ = 0.54967

8) The expected pressure is the conditional mean given by

$E(Y\mid x) = K\int\limits^{30}_{20} yh(y \mid x)\, dy$

$E(Y\mid x) = K\int\limits^{30}_{20} 10.066y^3+6291.39y\, dy = \frac{3}{380000} \times 3208609.27153$

= 25.33 psi

The standard deviation is given by

$Standard \, deviation =\sqrt{Variance}$

Variance = $K\int\limits^{30}_{20} [y-E(Y\mid x) ]^2h(y \mid x)\, dy$

$=K\int\limits^{30}_{20} [y-25.33]^2(10.066y^2+6291.39)\, dy$

= $\frac{3}{380000} \times 1047259.78$ = 8.268

The standard deviation = √8.268 = 2.875.

3 0

2 years ago

33 points!! The line contains the point (4,-2) and is perpendicular to 4x+3y=-6 The equation of the line is:

Deffense [45]

Answer:

y = 3/4 x - 5

Step-by-step explanation:

The slopes of perpendicular lines are negative reciprocals. We find the slope of the given line. Then we can find the slope of the reciprocal.

4x + 3y = -6

Solve for y.

3y = -4x - 6

y = -4/3 x - 2

The slope of the original line is -4/3.

The slope of the perpendicular is 3/4.

Now we need to find the equation of line with slope 3/4 that contains point (4, -2).

y = mx + b

-2 = (3/4)(4) + b

-2 = 3 + b

-5 = b

b = -5

The equation is

y = 3/4 x - 5

5 0

2 years ago

1. Use the Fermi process to estimate the number of bricks needed to fill an empty bathtub. Show your work.

exis [7]

According to sources, the most probable answer to this query are:
1) The volume of the rectangular prism given with the dimensions is 64 inches (lxwxh)
2) The volume of the rectangular prism given the dimensions is 32400 inches

Thank you for your question. Please don't hesitate to ask in Brainly your queries.

3 0

2 years ago

Read 2 more answers

Calculus application

Ugo [173]

Answer:

49m/s

59.07 m/s

Step-by-step explanation:

Given that :

Distance (s) = 178 m

Acceleration due to gravity (a) = g(downward) = 9.8m/s²

Velocity (V) after 5 seconds ;

The initial velocity (u) = 0

Using the relation :

v = u + at

Where ; t = Time = 5 seconds ; a = 9.8m/s²

v = 0 + 9.8(5)

v = 0 + 49

V = 49 m/s

Hence, velocity after 5 seconds = 49m/s

b) How fast is the ball traveling when it hits the ground?

V² = u² + 2as

Where s = height = 178m

V² = 0 + 2(9.8)(178)

V² = 0 + 3488.8

V² = 3488.8

V = √3488.8

V = 59.07 m/s

8 0

2 years ago

Manuel has 5/8 of a pizza to share equally with his brother. What fraction of the original pizza will each person receive?

Irina-Kira [14]

Answer:

1¼

Step-by-step explanation:

⅝ ÷½

=⅝×²/1

=⁵/4

=1¼

Whenever you divide two fractions you rewrite it switching the division sign for multiplication and writing the reciprocal of the second fraction.

3 0

3 years ago