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Mariulka [41]
3 years ago
8

Because of the increase in traffic between Springfield and Orangeville, a new road was built to connect the two towns. The old r

oad goes south x miles from Springfield to Freeport and then goes east x + 5 miles from Freeport to Orangeville. The new road is 9 miles long and goes straight from Springfield to Orangeville. Find the number of miles that a person drove using the old road. Please help ASAP!!! :(
Mathematics
1 answer:
Doss [256]3 years ago
5 0
<h3>Answer:</h3>

11.70 miles

<h3>Explanation:</h3>

The road from Springfield to Orangeville is apparently the hypotenuse of a right triangle whose legs are x and x+5. The value of x can be found from the Pythagorean theorem:

... 9² = x² + (x+5)²

... 2x² +10x -56 = 0 . . . . . . eliminate parentheses, collect terms, subtract 81.

... x² +5x -28 = 0 . . . . . . . . divide by 2 (because we can)

... x² +5x +6.25 = 28 +6.25 . . . . add 28, then add (5/2)² to complete the square

... (x +5/2) = √34.25 . . . . . . . .take the positive square root

... x = -2.5 +√34.25 ≈ 3.352

The total distance via the old road is the sum of the leg lengths:

... x + (x+5) = 2x +5 = 2(3.352) +5 = 11.704 . . . . miles

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Step-by-step explanation:

Three consecutive odd numbers: x, x+2, x+4 (since odd numbers occur every other number)

Just to show you that the numbers are x, x+2, and x+4, just say x is 13 then x+2 is 15 and x+4 is 17 (3 consecutive odd numbers)

Now the second equation:

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Step-by-step explanation:

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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
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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

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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

So the value of E(x,y) is 36

Part 9

The value is given as

Cov(X,Y)=E(x,y)-E(x)E(y)\\Cov(X,Y)=36-(3.6667)(4.6667)\\Cov(X,Y)=18.8886\\

So the value of Cov(x,y) is 18.8886

Part 10

The variables X and Y are considered independent when

f_{xy}(x,y)=f_x(x).f_y(y)\\

Here

f_x(x).f_y(y)=\frac{x-1}{8}\frac{10-y}{32} \\

And

f_{xy}(x,y)=\frac{1}{16}

As these two values are not equal, this indicates that X and Y are not independent variables.

4 0
3 years ago
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