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2 years ago

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Suppose we are given 4 sets A, B, C, D such that A ⊆ B and C ⊆ D such that A and C have no elements in common. Prove or give a c

ounterexample to the assertion that A and D also have no elements in common.

1 answer:

Mekhanik [1.2K]2 years ago

8 0

Here's a counterexample: let

$B = \{1, 2, 3, 4, 5\},\quad D = \{A, B, C, D, 5\}$

We choose the subsets as follows:

$A = \{1, 5\},\quad C = \{A, B, C\}$

It is true that $A\subseteq B$ and $C\subseteq D$ and that $A\cap C=\emptyset$ , but $A\cap D = \{5\}$

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Suppose you want to assess student attitudes about the new campus center by surveying 100 students at your school. In this examp

sleet_krkn [62]

Answer:

Suppose you want to assess student attitudes about the new campus center by surveying 100 students at your school. In this example, the group of 100 students represents the Sample, and all of the students at your school represent the Population.

Step-by-step explanation:

Previous concepts

The term sample represent a set of observations or individuals selected from a population. And we can have a random sample (when all the individuals have the same probability of being selected) or a non random sample (when not all the individuals have probability of inclusion into the sample)

The term population represent the total of observations or individuals with a common characteristic.

If N represent the sample of the population and n the sample size we have always this inequality:

$n \leq N$

Solution to the problem

Suppose you want to assess student attitudes about the new campus center by surveying 100 students at your school. In this example, the group of 100 students represents the Sample, and all of the students at your school represent the Population.

3 0

2 years ago

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

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

8 0

2 years ago

A worker has a base pay of $20 per hour, gets $3 more per hour after 8 hours, and gets $4 more

GalinKa [24]

Answer:

177 $

Step-by-step explanation:

20 x8 = 160

3x3= 9

4x2=8

8+9+160= 177

6 0

2 years ago

Orlando bought a new couch for $2,904, using the furniture store's finance plan. He will pay $121 a month for 24 months. Which e

jolli1 [7]

Answer : y=<span>2,904-121x , x for month

when you replace any number in "x" you mean how many months you pay (witch mean the sums of the all (121$) ) so when you </span>Subtract this all 121$ value you will get the the value of what you <span>still owes </span>

4 0

2 years ago

Anne buys her store's art supplies from a wholesaler. She purchases 8 canvases that are 16 by 20 inches for $7.16

Sladkaya [172]

Answer:

OC. $57.28

Step-by-step explanation:

It depends on how much she sells them for.

Let's say Anne sells them for $x.

Then her profit is:

8x-(8*7.16)

=8x-57.28

To make a minimum to at least make even is $57.28 and sell for $7.16 a piece.

7 0

2 years ago

Read 2 more answers