What is the domain of the function y=2.X-6?<br>

Samuel can do 120 jumping jacks in two minutes.

Natali5045456 [20]

Answer:

The ratio of jumping jacks to minutes is 120:2 or 60:1. He can do 60 Jumping jacks in 1 minute

Step-by-step explanation:

3 0

3 years ago

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The sum of three consecutive integers is 90. Find the three numbers by letting x

Law Incorporation [45]

Answer:

x = 28

Step-by-step explanation:

let us take three consecutive numbers be (x + 1), (x + 2) and (x + 3)

Now,

The sum of three consecutive integers is 90

i.e.,

(x + 1) + (x + 2) + (x + 3) = 90

x + 1 + x + 2 x + 3 = 90

3x + 6 = 90

3x = 90 - 6

3x = 84

x = 84/3

x = 28

x = 28Thus, The value of x is 28

<h2><em><u>VERIFICATION</u></em><em><u>:</u></em></h2>

(x + 1) + (x + 2) + (x + 3) = 90

(28 + 1) + (28 + 2) + (28 + 3) = 90

28 + 1 + 28 + 2 + 28 + 3 = 90

90 = 90

8 0

3 years ago

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What is the distance between (12, -7) and (-3, -7)?

gladu [14]

Answer:15 units

Step-by-step explanation:

3 0

3 years ago

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Which of the following is a trinomial with a constant term?

Pepsi [2]

Answer:

is A

Step-by-step explanation:

4 0

3 years ago

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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

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

5 0

3 years ago

What is the domain of the function y=2.X-6? - 0 O 0 O 3 6