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3 years ago

Is 1/3 a rational number

Mathematics

2 answers:

NNADVOKAT [17]3 years ago

3 0

Answer:

Yes 1/3 is a rational number.

Explanation:

in mathematics, a rational number is a number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. Hence, 1/3 is a rational number.

Ilia_Sergeevich [38]3 years ago

3 0

Answer:

Yes, it is a rational number.

Step-by-step explanation:

A rational number is any number that can be expressed as a ratio of integers. Even a fraction in which both the numerators and denominators are both integers but the denominator can never ever be less than 0.

I hope this helped you!

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Panel of judges was asked to judge the quality of different kinds of potato chips. The scatterplot and regression line below sho

nika2105 [10]

Answer:

Step-by-step explanation:

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6 0

4 years ago

Will mark Brainliest if someone can help me!!!!! (Idk if Prinicipal is right 2-5)

Talja [164]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$1500\\ r=rate\to r\%\to \frac{r}{100}\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1,2,3,4,5 \end{cases}$

$\bf A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 1}\implies A=1500(1.04)^1\implies A\approx 1560\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 2}\implies A=1500(1.04)^2\implies A\approx 1622.4\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 3}\implies A=1500(1.04)^3\implies A\approx 1687.3\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 4}\implies A=1500(1.04)^4\implies A\approx 1754.79\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 5}\implies A=1500(1.04)^5\implies A\approx 1824.98$

6 0

3 years ago

2) X and Y are jointly continuous with joint pdf

Nady [450]

From what I gather from your latest comments, the PDF is given to be

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}$

and in particular, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)

(a) Find c such that f is a proper density function. This would require

$\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}$

(b) Get the marginal density of X by integrating the joint density with respect to y :

$f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(d) The conditional distribution of X given Y can obtained by dividing the joint density by the marginal density of Y (which follows directly from the definition of conditional probability):

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

(e) From the definition of expectation:

$E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}$

$E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}$

$E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}$

(f) Note that the density of X | Y in part (d) identical to the marginal density of X found in (b), so yes, X and Y are indeed independent.

The result in (e) agrees with this conclusion, since E[XY] = E[X] E[Y] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)

8 0

4 years ago

Please please please help

irga5000 [103]

3x+x= 180°(straight angle)

4x= 180°

x= 180/4

x= 45°

5 0

3 years ago

Read 2 more answers

A student bought 3 boxes of pencils and 2 boxes of pens for $6. He then bought 2 boxes of pencils and 4 boxes of pens for $8. Fi

Annette [7]

Let c and n represent the numbers of pencil and pen boxes, respectively
.. 3c +2n = 6.00
.. 2c +4n = 8.00

You can halve the second equation and subtract it from the first to get
.. (3c +2n) -(c +2n) = 6.00 -4.00
.. 2c = 2.00
.. c = 1.00

Then
.. 1.00 +2n = 4.00 . . . . . half the original second equation with c filled n
.. 2n = 3.00
.. n = 1.50

A box of pencils costs $1.00; a box of pens costs $1.50.

8 0

4 years ago

Read 2 more answers

Is 1/3 a rational number​

Is 1/3 a rational number