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3 years ago

What is the difference between an irrational number and an integer? PLEASE NEED HELP ASAP!!!!

2 answers:

tatiyna3 years ago

4 0

Irrational numbers are numbers that have decimals and the decimals randomly keep going on. Remember, a repeating decimal is not necessarily an irrational number. (pi, sqrt(2), etc)
An integer is a whole number. (1, 2, 3, 4, etc) it could also be negative.

Tanzania [10]3 years ago

3 0

An irrational number is any real number that cannot be expressed as a ratio a/b, where a and b are integers, with b non-zero, and is therefore not a rational number. An integer is any real whole number. Basically, this means that an irrational number cannot be represented as a simple fraction.

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What is 10^{2} /19^{4}

MatroZZZ [7]

Answer:

$\frac{100}{130321}$

Step-by-step explanation:

10² ÷ $19^{4}$

100 ÷ $19^{4}$

100 ÷ 130321

$\frac{100}{130321}$

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3 years ago

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Dennis_Churaev [7]

Answer:

10648

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3 years ago

To the nearest tenth what is the length of the hypotenuse of a right triangle if one side measures 4.5meters and the other side

allsm [11]

Answer:

Step-by-step explanation:

c^2 = a^2 + b^2

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3 0

3 years ago

Suppose that the joint p.d.f. of a pair of random variables (X, Y ) is constant on the rectangle where 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1,

kati45 [8]

Answer:

A. p.d.f(x,y)= 1/2

B. P(X>Y)= 3/4

Step-by-step explanation:

A. We are told that p.d.f(x,y) is constant on the rectangle 0<x<2, 0<y<1. To find that constant the property of probabilities that is useful here is the following:

$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}p.d.f(x,y)\ dx\ dy=1$

We know that f(x,y)=c (c is a constant) in the rectangle and outside of it f(x,y)=0, so the integral above could be rewritten easily as:

$\int_{0}^{2}\int_{0}^{1}c\ dy\ dx=1$

Solving the integral it´s possible to find the value c:

$\int_{0}^{2}\int_{0}^{1}c\ dy\ dx=\int_{0}^{2}c\ dx$

$\int_{0}^{2}c\ dx= 2c$

$2c=1\ \rightarrow c=1/2$

This is the value of f(x,y) in the rectangle

B. Now that we know our p.d.f(x,y), we are asked to find P(X>Y) in the rectangle. There´s a graph of the area (probability) that we are looking for, it´ll make easy the understanding of the move that we are going to make.

We are asked for the probability that X>Y, i.e the part of the rectangle below the graph of X=Y. Because x is between 0 and 2, whenever x>1, it will immediately be greater than y. If x<1, we just need to assure that 0<y<x. With this data we can find the probability this way:

$P(X>Y)=\int_{0}^{1}\int_{0}^{x}1/2\ dy\ dx + \int_{1}^{2}\int_{0}^{1}1/2\ dy\ dx$

The first integral represents the area (that´s equal to the probability) of the triangle at the left of the blue line, and the other integral it´s related to the square that´s right of the blue line.

And now we are ready to solve the problem:

$P(X>Y)=\int_{0}^{1}\frac{x}{2}\ dx + \int_{1}^{2}1/2\ dx$

$P(X>Y)=\frac{x^{2}}{4}|_{0}^{1} + 1/2$

$P(X>Y)=1/4 + 1/2$

We conclude that P(X>Y)=3/4

8 0

4 years ago