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2 years ago

Line___is a transversal.

Mathematics

2 answers:

tester [92]2 years ago

6 0

Answer:

line C

Step-by-step explanation:

iVinArrow [24]2 years ago

4 0

Answer:

Step-by-step explanation:

Line A is Parallel to Line B

Line C intersects Line A and B

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2 years ago

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2. Sam is climbing 2/3 meters every 1/3 of an hour. How many meters does

ICE Princess25 [194]

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2/9

Step-by-step explanation:

if u multiply 2/3 × 1/3 u get 2/9

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3 years ago

Write two different mixed numbers where the LCD (lowest common denominator) is 12.<br> HELP ASAP

alex41 [277]

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1 1/2, 3 1/3

<h3>Step-by-step explanation:</h3>

For the denominators (3, 4) the least common multiple (LCM) is 12. Calculations to rewrite the original inputs as equivalent fractions with the LCD:

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3 years ago

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2 years ago

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Let X represent the amount of energy a city uses (in megawatt-hours) in the Kanto region. Let Y represent the amount of mismanag

Setler [38]

Answer:

Part 2: The probability of X≤2 or X≥4 is 0.5.

Part 3: The value of marginal probability of y is $f_y(y)=\frac{10-y}{32}$ for $2\leq y\leq 10$

Part 4:The value of E(y) is 4.6667.

Part 5:The value of $f_{xy}(x)$ is $\frac{2}{10-y}$ for $2\leq y\leq 2x\leq 10$

Part 6:The value of $M_{x,y}(y)$ is $\frac{y+10}{4}$

Part 7:The value of E(x) is 3.6667.

Part 8:The value of E(x,y) is 36.

Part 9:The value of Cov(x,y) is 18.8886.

Part 10:X and Y are not independent variables as $f_{xy}(x,y)\neq f_x(x).f_y(y)\\$

Step-by-step explanation:

As the complete question is here, however some of the values are not readable, thus the question is found online and is attached herewith.

From the given data, the joint distribution is given as

$f(x,y)=\frac{1}{16}$ for $2\leq y\leq 2x\leq 10$

Now the distribution of x is given as

$f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy$

Here the limits for y are $2\leq y\leq 2x$ So the equation becomes

$f_x(x)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dy\\f_x(x)=\int\limits^{2x}_{2} \frac{1}{16} \, dy\\f_x(x)=\frac{1}{16} (2x-2)\\f_x(x)=\frac{x-1}{8} \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 1\leq x\leq 5$

Part 2:

The probability is given as

$P(X\leq 2 U X\geq 4)=\int\limits^2_1 {f_x(x)} \, dx +\int\limits^5_4 {f_x(x)} \, dx\\P(X\leq 2 U X\geq 4)=\int\limits^2_1 {\frac{x-1}{8}} \, dx +\int\limits^5_4 {\frac{x-1}{8}} \, dx\\P(X\leq 2 U X\geq 4)=\frac{1}{16}+\frac{7}{16}\\P(X\leq 2 U X\geq 4)=0.5$

So the probability of X≤2 or X≥4 is 0.5.

Part 3:

The distribution of y is given as

$f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx$

Here the limits for x are $y/2\leq x\leq 5$ So the equation becomes

$f_y(y)=\int\limits^{\infty}_{-\infty} {f(x,y)} \, dx\\f_y(y)=\int\limits^{5}_{y/2} \frac{1}{16} \, dx\\f_y(y)=\frac{1}{16} (5-\frac{y}{2})\\f_y(y)=\frac{10-y}{32} \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 2\leq y\leq 10$

So the value of marginal probability of y is $f_y(y)=\frac{10-y}{32}$ for $2\leq y\leq 10$

Part 4

The value is given as

$E(y)=\int\limits^{10}_2 {yf_y(y)} \, dy\\E(y)=\int\limits^{10}_2 {y\frac{10-y}{32}} \, dy\\E(y)=\frac{1}{32}\int\limits^{10}_2 {10y-y^2} \, dy\\E(y)=4.6667$

So the value of E(y) is 4.6667.

Part 5

This is given as

$f_{xy}(x)=\frac{f_{xy}(x,y)}{f_y(y)}\\f_{xy}(x)=\frac{\frac{1}{16}}{\frac{10-y}{32}}\\f_{xy}(x)=\frac{2}{10-y}$

So the value of $f_{xy}(x)$ is $\frac{2}{10-y}$ for $2\leq y\leq 2x\leq 10$

Part 6

The value is given as

$\geq M_{x,y}(y)=E(f_{xy}(x))=\int\limits^5_{y/2} {x f_{xy}(x)} \, dx \\M_{x,y}(y)=\int\limits^5_{y/2} {x \frac{2}{10-y}} \, dx \\M_{x,y}(y)=\frac{2}{10-y}\left[\frac{x^2}{2}\right]^5_{\frac{y}{2}}\\M_{x,y}(y)=\frac{2}{10-y}\left(\frac{25}{2}-\frac{y^2}{8}\right)\\M_{x,y}(y)=\frac{y+10}{4}$

So the value of $M_{x,y}(y)$ is $\frac{y+10}{4}$

Part 7

The value is given as

$E(x)=\int\limits^{5}_1 {xf_x(x)} \, dx\\E(x)=\int\limits^{5}_1 {x\frac{x-1}{8}} \, dx\\E(x)=\frac{1}{8}\left(\frac{124}{3}-12\right)\\E(x)=\frac{11}{3} =3.6667$

So the value of E(x) is 3.6667.

Part 8

The value is given as

$E(x,y)=\int\limits^{5}_1 \int\limits^{10}_2 {xyf_{x,y}(x,y)} \,dy\, dx\\E(x,y)=\int\limits^{5}_1 \int\limits^{10}_2 {xy\frac{1}{16}} \,dy\, dx\\E(x,y)=\int\limits^{5}_1 \frac{x}{16}\left[\frac{y^2}{2}\right]^{10}_2\, dx\\E(x,y)=\int\limits^{5}_1 3x\, dx\\\\E(x,y)=3\left[\frac{x^2}{2}\right]^5_1\\E(x,y)=36$