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lianna [129]
3 years ago
7

When multiplying a number by 10, move the decimal to the right. When multiplying a number by 0.1, move the decimal to the left.

Why? :( HELP IM TiMED
Mathematics
2 answers:
ki77a [65]3 years ago
6 0

Answer: 0.01 x 10 = .1

it moved ten spaces up or in simpler terms the decimal move one space to the right because the number is getting bigger.

Step-by-step explanation:

jek_recluse [69]3 years ago
6 0

Answer: When you multiply decimals they should add up otherwise move to the left but if you multiply by 10 the decimal moves to the right because of the change in numbers from ones to hundreds and so on. An easier way to unterstand is when you multiply the regular way don't multiply the decimal but count however many places that decimal takes up add it with the other decimals and the put it in the solution. EX:

5                      1.2

x.1                  x 2.28

--------         --------------

.5                     2.736

Step-by-step explanation:

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Answer:

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<u>Pre-Algebra</u>

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<u>Algebra I</u>

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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 &lt; x &lt; 1, 0 &lt; y &lt; 2
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I'm going to assume the joint density function is

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\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

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  • 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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