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3 years ago

An irrational number between 1/2 and 3/4

Mathematics

1 answer:

prohojiy [21]3 years ago

4 0

Answer:

2.743/4

I guess?

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If b/a=c/b=d/c=2/3, find the value of a:b:c:d

leva [86]

Answer:

the ratio would be 6.75:4.5:3:2

Step-by-step explanation:

replace d and c with

d=2 and c=3

Then using proportion multiplication, you will find that b is 4.5

now replace all b with 4.5.

Then again using proportion multiplication, you will get a is 6.75.

Plug in a, b, c, d in the ratio and you get 6.75:4.5:3:2

5 0

2 years ago

An equation in the system is y = x + 1. What is the other equation in the system if the solution is only at (0, 1)?

Setler [38]

<span>x + y = 0 hope this helps </span>

6 0

3 years ago

Which point is on the circle centered at the origin with a radius of 5 units?

Aleonysh [2.5K]

Answer:

(2,√21)

Step-by-step explanation:

The circle centered at the origin has equation:

${x}^{2} + {y}^{2} = 25$

Any point that satisfies this equation lie on this circle.

When x=2, we substitute and solve for y.

${2}^{2} + {y}^{2} = 25 \\ 4 + {y}^{2} = 25 \\ {y}^{2} = 25 - 4 \\ {y}^{2} = 21$

Take square root to get:

$y = \pm \sqrt{21}$

Therefore (2,-√21) and (2,√21) are on this circle.

From the options, (2,√21) is the correct answer

3 0

3 years ago

Read 2 more answers

a crane lifts a pallet of concrete blocks 9 ft from the back of the truck. the truck drives away and the crane lowers the pallet

goblinko [34]

Answer:
The final position is 5 feet below the back of the truck
Step-by-step explanation:
* Lets explain how to solve the problem
- A crane lifts a pallet of concrete blocks 8 feet from the back of
a truck
- The crane lowers the pallet 13 feet after the truck drives away
- Assume that the zero level of the position of the ballet of concrete
blocks is the back of the truck
∵ The crane lifts the pallet of concrete blocks 8 feet from the back
of the truck
- That means it take the pallet from zero to 8
∴ The height of the pallet of concrete blocks is 8 feet over
the starting position
∵ The crane lowers the pallet of concrete blocks 13 feet
- That means the crane lower the pallet from the height 8 and
lift it down 13 feet, so we must to take out from the 8 feet the
13 feet to find the final position of the pallet of concrete blocks
∴ The pallet position is ⇒ 8 - 13 = -5
∴ The position of the pallet of concrete blocks is 5 feet below the
starting position which is the back of the truck
* The final position is 5 feet below the back of the truck

8 0

3 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