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2 years ago

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Please helppp!!! Lasandra wants to center a towel bar on her door that is 29 inches wide she determines that the distance from e

ach end of the towel bar to the end of the door is 7.75 inches write and solve an equation to find the length of the towel bar

1 answer:

lapo4ka [179]2 years ago

8 0

Answer: 13.5 inches

Step-by-step explanation:

Let x be the length of the towel bar.

Since, the towel bar's ends from each end of the door are 7.75 each,

Thus, the width of the door = x + 7.75 + 7.75 = x + 15.5

But, According to the question,

The width of the door = 29 inches,

Therefore, x + 15.5 = 29

⇒ x = 13.5 inches

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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

exis [7]

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

Repair calls are handled by one repairman at a photocopy shop. Repair time, including travel time, is exponentially distributed,

seropon [69]

Answer and Step-by-step explanation:

Data provided in the question

Mean = 1.1 hours per call =

R = Mean rate = 1.6 per eight hour day

$\mu$ = $\frac{8}{1.6}$ = 5 per day

Based on the above information

a. The average number of customers is

$= \frac{R^2}{\mu(\mu- R)}$

$= \frac{1.6^2}{5(5- 1.6)}$

= 151

b. The system utilization is

$= \frac{R}{\mu}$

= $\frac{1.6}{5}$

= 0.32

c. The amount of time required is

= 1 - system utilization

= 1 - 0.32

= 0.68

And, there is 8 hours per day

So, it would be

= $0.68 \times 8$

= 5.44 hours

d. Now the probability of two or more customers is

$= 1 - (0.68 + 0.68 \times 0.32)$

= 0.1024

Therefore we simply applied the above formulas

4 0

2 years ago

The distribution of the amount of money spent by students for textbooks in a semester is approximately normal in shape with a me

kkurt [141]

The final part of the question is asking;

How much did all (99.7%) of the students spend on textbooks in a semester

Answer:

almost all (99.7%) of the students spent between $165 and $315 on textbooks in a semester.

Step-by-step explanation:

The standard deviation rule describes to us that for distributions that have the normal shape, approximately 99.7% of the observations fall within 3 standard deviations of the mean.

In this question, we are given that; Mean = 240 and Standard deviation= 25

So, 3 standard deviation below the mean = Mean - 3(standard deviation)

= 240 - (3 × 25)

= 240 - 75 = 165

Now, 3 standard deviation above the mean = Mean + 3 standard deviation = 240 + (3 × 25)

= 240 + 75 = 315

So, almost all (99.7%) of the students spent between $165 and $315 on textbooks in a semester.

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2 years ago

A local library claims it has 2.73x103 books. If there are 42 people in the building, how many books per capita are there?

sveticcg [70]

Answer:

$Per\ Capita = 64$

Step-by-step explanation:

Given

$Books = 2.7 * 10^3$

$People = 42$

Required

Determine the per capita

The per capita is calculated by dividing number of books by people as follows;

$Per\ Capita = \frac{Books}{People}$

Substitute 2.7 * 10^3 for books and 42 for people

$Per\ Capita = \frac{2.7 * 10^3}{42}$

$Per\ Capita = 64$ (Approximated)

<em>Hence, the per capita is 64</em>

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2 years ago

If the trapezoid below is reflected across the x-axis, what are the coordinates of B'?

Grace [21]

The x-coordinate remains the same as the x-coordinate of point B.
The y-coordinate becomes the additive inverse of the y-coordinate of point B.

Answer: B. (3, -8)

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2 years ago

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