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2 years ago

Please help.... asap

Mathematics

1 answer:

DaniilM [7]2 years ago

6 0

Answer:

GCF is less than 3:

- 12 and 13

GCF is equal to 3:

- 6 and 27

- 12 and 21

- 3 and 6

GCF is greater than 3:

- 4 and 12

- 6 and 18

Step-by-step explanation:

6 and 27 GCF: 3

12 and 13 GCF: 1

4 and 12 GCF: 4

3 and 6 GCF: 3

12 and 21 GCF: 3

6 and 18 GCF: 6

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The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

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3 years ago

Which table of values represents a linear function?

krek1111 [17]

Step-by-step explanation:

I'm learn this too hahaha

6 0

3 years ago

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What is the value of the expression 24 + 10•8?

Akimi4 [234]

Answer:

$\large\boxed{\text{The expression is equal to 104}}$

Step-by-step explanation:

$24+10 \cdot 8 \\24+80\\104$

4 0

3 years ago

What is the value of the expression?<br> 3⋅[(30−8)÷2+2]

dusya [7]

Answer: 39

Step-by-step explanation:

30-8=22

22/2=11

11+2=13

13 times three is (39)

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3 years ago

Mathematical induction of 3k-1 ≥ 4k ( 3k = k power of 3 )

tresset_1 [31]

For making mathematical induction, we need:

a base case

An $n_0$ for which the relation holds true

the induction step

if its true for $n_i$ , then, is true for $n_{i+1}$

the relationship is not true for 1 or 2

$1^3-1 = 0 < 4*1$

$2^3-1 = 8 -1 = 7 < 4*2 = 8$

but, is true for 3

$3^3-1 = 27 -1 = 26 > 4*3 = 12$

<h3>induction step</h3>

lets say that the relationship is true for n, this is

$n^3 -1 \ge 4 n$

lets add 4 on each side, this is

$n^3 -1 + 4 \ge 4 n + 4$

$n^3 + 3 \ge 4 (n + 1)$

now

$(n+1)^3 = n^3 +3 n^2 + 3 n + 1$

$(n+1)^3 \ge n^3 + 3 n$

if $n \ge 1$ then $3 n \ge 3$ , so

$(n+1)^3 \ge n^3 + 3 n \ge n^3 + 3$

$(n+1)^3 \ge n^3 + 3 \ge 4 (n + 1)$

$(n+1)^3 \ge 4 (n + 1)$

and this is what we were looking for!

So, for any natural equal or greater than 3, the relationship is true.

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3 years ago