NEED HELP PLZZZZxxxx

Question is attached

Arlecino [84]

Answer:

Let v = ml of 100% vinegar

Then 150-v = ml of dressing

v + .05(150-v) = .24(150)

v + 7.5 - .05v = 36

.95v = 28.5

v = 28.5/.95

v = 30 ml of vinegar

dressing = 150-30 = 120 ml

Step-by-step explanation:

7 0

3 years ago

What is the autput the following machine when the intut is 4

ch4aika [34]

Input is 4.

Process machine:

Input > - 7 > ÷ 3 > Output

Solve:

(Input) 4 - 7 = -3

-3 ÷ 3 = -1 (Output)

Input = 4

Output = -1

3 0

3 years ago

The expanded form of 6,398 is

masya89 [10]

Answer:

The expanded form of 6,398 is 6000 + 300 + 90 + 8

3 0

3 years ago

Read 2 more answers

JJill begins the summer with $500 in her savings account. Each week, she withdraws $20. Her brother, Sam,starts the summer with

meriva

Answer:

7th week

Step-by-step explanation:

Let us represent the number of weeks as x

Jill begins the summer with $500 in her savings account. Each week, she withdraws $20.

$500 - $20× x

500 - 20x

Her brother, Sam,starts the summer with $150 in his account. He adds $30 to his account each week.

$150 + $30 × x

150 + 30x

At what week will both Jill and Sam have the same amount in their account?

We solve this be Equating both Equations together

Jill = Sam

500 - 20x = 150 + 30x

Collect like terms

500 - 150 = 30x + 20x

350 = 50x

Divide both sides by 50

50x/50 = 350/50

x = 7 weeks

Therefore, at the 7 week, Jill and sam would have the same amount in their accounts

3 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

4 years ago