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2 years ago

-4(a-12) help me because I am really confused

Mathematics

2 answers:

seraphim [82]2 years ago

7 0

Answer:

-4a+48

Step-by-step explanation:

Multiply -4 to a and -12

-4a+48

White raven [17]2 years ago

5 0

Answer:

-4a+48

Step-by-step explanation:

Using the distributive property, you would multiply -4 with the values inside the brackets, which are a and -12.

-4*a + (-4*(-12))

= -4a+(48)

= -4a+48

Hope this helps!

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A cake shop bakes a variety of brownies. the top-selling brownies are ones with toppings of chocolate chip, walnuts, or both. a

sesenic [268]

The probability that they will pick neither the chocolate chip nor the walnut toppings is 0.3

We have

the total of all probabilities is 1.00, or 100%.

In the Venn diagram, we have the probabilities 0.2, 0.4 and 0.1; these sum to

0.2+0.4+0.1 = 0.6+0.1 = 0.7.

This leaves us 1.00-0.7 = 0.3 for the remaining probability of no toppings.

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6 0

1 year ago

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

ANWSER both questions pls for brainly

hjlf

Answer:

40x-24

3b(6c-3a+c)

Step-by-step explanation:

For the first one use distributive property and multiply each term by 8.

For the second one find a common factor which is 3b and divide all terms by that number.

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2 years ago

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Kipish [7]

Which one has the variable?

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3 years ago

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Airida [17]

Answer:

Step-by-step explanation:

Hey There so the first step of this problem is to find the formula for circumference

$C=2\pi r$

given some of the values we can form an equation to solve for the radius (r)

$31.4=3.14(2)r\\3.14(2)=6.28\\31.4=6.28r\\\frac{31.4}{6.28} =5$

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3 years ago

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