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Lubov Fominskaja [6]
3 years ago
12

Find x(g) for the graph.

Mathematics
1 answer:
DanielleElmas [232]3 years ago
3 0
The answer you are looking for AC3
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Which of the following is equivalent to 5 · (7 · 4)? *
MariettaO [177]
B because you multiply 7 times 4 that’s 28 then times 5 that’s 140 so for b you do 5 times 7 that’s 35 then times 4 that’s 140 so b is the correct answer, hope it helps
8 0
2 years ago
Camila was out at a restaurant for dinner when the bill came. Her dinner came to $32. After adding in a tip, before tax, she pai
Gnesinka [82]

Answer:

25%

Cause 125%-100%=25%

​

3 0
2 years ago
A snail travels 12 centimeters in 4 minutes. The snail travels the same distance each minute. How many milliliters did the snail
Dmitry_Shevchenko [17]

Answer:

the snail travels 30 mm per minute

Step-by-step explanation:

Well 12 cenitmters divided by 4 minutes equals the amount of centimeters the snail travels per minute, which is 3 centimeters (12 / 4 = 3). but be careful cuz we need to convert the 3 cm to mm. 1 cm = 10 mm so 3 cm = 30 mm.

the snail travels 30 mm per minute

6 0
2 years 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
Rhombus RSTU has vertices R(-1, 6), S(3, 7), T(2, 3), and U(-2, 2). If it's translated under the rule (x, y) LaTeX: \longrightar
Andreas93 [3]
(4,3) you add 5 to the x and subtract 3 from y
7 0
3 years ago
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