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scoundrel [369]

3 years ago

15

Mr. Cannon has a piece of land that is 3/7 of an acre. He wants to build 10 houses on his land. How much land does each house ge

t?
Show me the division statement
explain why you chose the dividend and the divisor
give me the quotient.

1 answer:

sashaice [31]3 years ago

4 0

Answer:

Step-by-step explanation:

3/7 divided by 10 houses= 0.042 acres per house

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Compute 33(0.5) - 6(0.5) - 7/2.

lora16 [44]

Hi there! The answer is 10.

$33(0.5) - 6(0.5) - \frac{7}{2}$
We can use PEMDAS (Parenthesis, exponents, multiply, divide, add, subtract) to find our answer.

Since we cannot work inside the parenthesis in this question and since we don't have exponents, we can move on to multiplying.
$16.5 - 3 - \frac{7}{2}$

Our next step would be subtracting. To make the process of subtracting some easier, I'll first change the fraction into a decimal.
$16.5 - 3 - 3.5$

Finally subtract.
$13.5 - 3.5 = 10$

The answer is 10.
~ Hope this helps you!

5 0

3 years ago

Read 2 more answers

Simplify 2/5 + 1/3<br><br> I NEED THIS ASAP

lana66690 [7]

Answer:

(11/15)'s

Step-by-step explanation:

(2/5) is equal to (6/15)

(1/3) is equal to (5/15)

(5/15)+(6/15)=(11/15)

5 0

3 years ago

Need help pls explain

Scilla [17]

Answer:

C. 93°

Step-by-step explanation:

WY is the diameter.

WX and XY sum to 180° as half the circle.

<u>Central angle is equal to intercepted arc:</u>

mXY = m∠XVY = 87°
mWX = 180° - mXY = 180° - 87° = 93°

Correct choice is C

4 0

3 years ago

Read 2 more answers

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

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

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3 years ago

What is the definition of a parent function?

Mice21 [21]

Answer:

Simón arre que si cuando quieras

8 0

3 years ago