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2 years ago

In ΔTUV, t = 410 cm, ∠U=27° and ∠V=78°. Find the length of v, to the nearest 10th of a centimeter.

Mathematics

1 answer:

telo118 [61]2 years ago

6 0

Answer:

$v \approx 415.2\,cm$

Step-by-step explanation:

The angle T is:

$\angle T = 180^{\circ} - 27^{\circ} - 78^{\circ}$

$\angle T = 75^{\circ}$

Now, the length of v is determine by the Law of Sines:

$\frac{410\,cm}{\sin 75^{\circ}} = \frac{v}{\sin 78^{\circ}}$

$v = (410\,cm)\cdot \left(\frac{\sin 78^{\circ}}{\sin 75^{\circ}} \right)$

$v \approx 415.2\,cm$

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Example 4.5 introduced the concept of time headway in traffic flow and proposed a particular distribution for X 5 the headway be

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Answer:

a. k = 3

b. Cumulative distribution function X, $F(x)=\left \{ {0} , x\leq 1 \atop {1-x^{-3}, x>1}} \right.$

c. Probability when headway exceeds 2 seconds = 0.125

Probability when headway is between 2 and 3 seconds = 0.088

d. Mean value of headway = 1.5

Standard deviation of headway = 0.866

e. Probability that headway is within 1 standard deviation of the mean value = 0.9245

Step-by-step explanation:

From the information provided,

Let X be the time headway between two randomly selected consecutive cars (sec).

The known distribution of time headway is,

$f(x) = \left \{ {\frac{k}{x^4} , x > 1} \atop {0} , x \leq 1 } \right.$

a. Value of k.

Since the distribution of X is a valid density function, the total area for density function is unity. That is,

$\int\limits^{\infty}_{-\infty} f(x)dx=1$

So, the equation becomes,

$\int\limits^{1}_{-\infty} f(x)dx + \int\limits^{\infty}_{1} f(x)dx=1\\0 + \int\limits^{\infty}_{1} {\frac{k}{x^4}}.dx=1\\0 + k \int\limits^{\infty}_{1} {\frac{1}{x^4}}.dx=1\\k[\frac{x^{-3}}{-3}]^{\infty}_1=1\\k[0-(\frac{1}{-3})]=1\\\frac{k}{3}=1\\k=3$

b. For this problem, the cumulative distribution function is defined as :

$F(x) = \int\limits^1_{\infty} f(x)dx + \int\limits^x_1 f(x)dx$

Now,

$F(x) = 0 + \int\limits^x_1 {\frac{k}{x^4}}.dx\\= 0 + \int\limits^x_1 3x^{-4}.dx\\= 3 \int\limits^x_1 x^{-4}dx\\= 3[\frac{x^{-4+1}}{-4+1}]^3_1\\= 3[\frac{x^{-3}}{-3}]^3_1\\=(\frac{-1}{x^3})|^x_1\\=(-\frac{1}{x^3}-(\frac{-1}{1}))=1- \frac{1}{x^3}=1-x^{-3}$

Therefore the cumulative distribution function X is,

$F(x)=\left \{ {0} , x\leq 1 \atop {1-x^{-3}, x>1}} \right.$

c. Probability when the headway exceeds 2 secs.

Using cdf in part b, the required probability is,

$P(X>2)=1-P(X\leq 2)\\=1-F(2)\\=1-[1-2^{-3}]\\=1-(1- \frac{1}{8})\\=\frac{1}{8} = 0.125$

Probability when headway is between 2 seconds and 3 seconds

Using the cdf in part b, the required probability is,

$P(2$

≅ 0.088

d. Mean value of headway,

$E(X)=\int\limits x * f(x)dx\\=\int\limits^{\infty}_1 x(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x(x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-3}dx\\=3[\frac{x^{-3+1}}{-3+1}]^{\infty}_1\\=3[\frac{x^{-2}}{-2}]^{\infty}_1\\=3[\frac{1}{-2x^2}]^{\infty}_1\\=3[- \frac{1}{2x^2}]^{\infty}_1\\=3[- \frac{1}{2(\infty)^2}- (- \frac{1}{2(1)^2})]\\=3(\frac{1}{2})=1.5$

And,

$E(X^2)= \int\limits^{\infty}_1 x^2(3x^{-4})dx\\=3 \int\limits^{\infty}_1 x^{-2} dx\\=3[- \frac{1}{x}]^{\infty}_1\\=3(- \frac{1}{\infty}+1)=3$

The standard deviation of headway is,

$= \sqrt{V(X)}\\ =\sqrt{E(X^2)-[E(X)]^2} \\=\sqrt{3-(1.5)^2} \\=0.8660254$

≅ 0.866

e. Probability that headway is within 1 standard deviation of the mean value

$P(\alpha - \beta < X < \alpha + \beta) = P(1.5-0.866 < X < 1.5 +0.866)\\=P(0.634 < X < 2.366)\\=P(X$

From part b, F(x) = 0, if x ≤ 1

$=1-(2.366)^{-3}\\=0.9245$

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2 years ago

Ramon wants to plant cucumbers and tomatoes in his garden. He has room for

yarga [219]

Answer:

Number of Cucumbers = 12

Number of Tomatoes = 4

Step-by-step explanation:

Let number of cucumber be c and number of tomatoes be t

Since he has room for 16 plants, we can write:

c + t = 16

He wants to plant 3 times as many cucumbers as tomatoes. We can write:

c = 3t

We can substitute this in 1st equation and solve for t:

c + t = 16

3t + t = 16

4t = 16

t = 16/4 = 4

And c = 3t

c = 3(4) = 12

Number of Cucumbers = 12

Number of Tomatoes = 4

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A mirror with a parabolic cross section is used to collect sunlight on a pipe located at the focus of the mirror. The pipe is lo

vodomira [7]

Answer: $x^2=32y$

Step-by-step explanation:

Given: A mirror with a parabolic cross section is used to collect sunlight on a pipe located at the focus of the mirror.

The pipe is located 8 inches from the vertex of the mirror.

Assume the vertex is at the origin.

If the parabola opens upwards

then the coordinates of focus= (0,8)

We know that equation of parabola with focus (0,a) and open upards is of the form (vertex=(0,0)) is

$x^2=4ay$

Substitute the value of a=8 in equation, we get

$x^2=4\times8y$

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