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kirill [66]
4 years ago
9

You represent a loss as a ______ amount positive negative

Mathematics
2 answers:
lesantik [10]4 years ago
6 0
Negative. If I had $10.00 and I have to give $11.00 then I would have -$1.00
Tamiku [17]4 years ago
3 0
Negative.
For example, if you had 1 apple, and subtract 2, you will get -1.
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Also this stuff too I need it before my teach put grades in
andrew11 [14]

Answer:

1. Distance is 5 units 2. Distance is 14 units 3. B. (2, -5), 10. (4, 14) 11. (6, 12) 14. x = 12 16. x = 55

Step-by-step explanation:

6 0
3 years ago
Someone plz help me <br> I will give brainliest
umka2103 [35]

Answer:

10 days

Step-by-step explanation:

2/.2=10

3 0
3 years ago
Read 2 more answers
Lakisha had $10 dollar bills and $1 bills in her wallet. She used seven bills to buy a pair of shoes for $43. How many of each t
daser333 [38]

Answer:

She spent 4 $10 bills and 3 $1 bills

Step-by-step explanation:

She used 7 bills and you have 3 $1 that makes the 3 of 43 then 4 $10 bills adds up to 40 so add 40 and 3 together you get the price of the shoes and if you add the 4 $10 bills and the 3 $1 bills you get 7

5 0
3 years ago
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Function P models the profit, P(x), a fitness company earns for making and selling x water bottles.
timurjin [86]

Answer:

The answer is 20 and 200.

Step-by-step explanation:

8 0
4 years ago
Let X and Y have the joint density f(x, y) = e −y , for 0 ≤ x ≤ y. (a) Find Cov(X, Y ) and the correlation of X and Y . (b) Find
adoni [48]

a. I assume the following definitions for covariance and correlation:

\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]

\mathrm{Corr}[X,Y]=\dfrac{\mathrm{Cov}[X,Y]}{\sqrt{\mathrm{Var}[X]\mathrm{Var}[Y]}}

Recall that

E[g(X,Y)]=\displaystyle\iint_{\Bbb R^2}g(x,y)f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy

where f_{X,Y} is the joint density, which allows us to easily compute the necessary expectations (a.k.a. first moments):

E[XY]=\displaystyle\int_0^\infty\int_0^yxye^{-y}\,\mathrm dx\,\mathrm dy=3

E[X]=\displaystyle\int_0^\infty\int_0^yxe^{-y}\,\mathrm dx\,\mathrm dy=1

E[Y]=\displaystyle\int_0^\infty\int_0^yye^{-y}\,\mathrm dx=2

Also, recall that the variance of a random variable X is defined by

\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2

We use the previous fact to find the second moments:

E[X^2]=\displaystyle\int_0^\infty\int_0^yx^2e^{-y}\,\mathrm dx\,\mathrm dy=2

E[Y^2]=\displaystyle\int_0^\infty\int_0^yy^2e^{-y}\,\mathrm dx\,\mathrm dy=6

Then the variances are

\mathrm{Var}[X]=2-1^2=1

\mathrm{Var}[Y]=6-2^2=2

Putting everything together, we find the covariance to be

\mathrm{Cov}[X,Y]=3-1\cdot2\implies\boxed{\mathrm{Cov}[X,Y]=1}

and the correlation to be

\mathrm{Corr}[X,Y]=\dfrac1{\sqrt{1\cdot2}}\implies\boxed{\mathrm{Corr}[X,Y]=\dfrac1{\sqrt2}}

b. To find the conditional expectations, first find the conditional densities. Recall that

f_{X,Y}=f_{X\mid Y}(x\mid y)f_Y(y)=f_{Y\mid X}(y\mid x)f_X(x)

where f_{X\mid Y} is the conditional density of X given Y, and f_X is the marginal density of X.

The law of total probability gives us a way to obtain the marginal densities:

f_X(x)=\displaystyle\int_x^\infty e^{-y}\,\mathrm dy=\begin{cases}e^{-x}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}

f_Y(y)=\displaystyle\int_0^ye^{-y}\,\mathrm dx=\begin{cases}ye^{-y}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}

Then it follows that the conditional densities are

f_{X\mid Y}(x\mid y)=\begin{cases}\frac1y&\text{for }0\le x

f_{Y\mid X}(y\mid x)=\begin{cases}e^{x-y}&\text{for }0\le x

Then the conditional expectations are

E[X\mid Y=y]=\displaystyle\int_0^y\frac xy\,\mathrm dy\implies\boxed{E[X\mid Y=y]=\frac y2}

E[Y\mid X=x]=\displaystyle\int_x^\infty ye^{x-y}\,\mathrm dy\implies\boxed{E[Y\mid X=x]=x+1}

c. I don't know which theorems are mentioned here, but it's probably safe to assume they are the laws of total expectation (LTE) and variance (LTV), which say

E[X]=E[E[X\mid Y]]

\mathrm{Var}[X]=E[\mathrm{Var}[X\mid Y]]+\mathrm{Var}[E[X\mid Y]]

We've found that E[X\mid Y]=\frac Y2 and E[Y\mid X]=X+1, so that by the LTE,

E[X]=E[E[X\mid Y]]=E\left[\dfrac Y2\right]\implies E[Y]=2E[X]

E[Y]=E[E[Y\mid X]]=E[X+1]\implies E[Y]=E[X]+1

\implies2E[X]=E[X]+1\implies\boxed{E[X]=1}

Next, we have

\mathrm{Var}[X\mid Y]=E[X^2\mid Y]-E[X\mid Y]^2=\dfrac{Y^2}3-\left(\dfrac Y2\right)^2\implies\mathrm{Var}[X\mid Y]=\dfrac{Y^2}{12}

where the second moment is computed via

E[X^2\mid Y=y]=\displaystyle\int_0^y\frac{x^2}y\,\mathrm dx=\frac{y^2}3

In turn, this gives

E\left[\dfrac{Y^2}{12}\right]=\displaystyle\int_0^\infty\int_0^y\frac{y^2e^{-y}}{12}\,\mathrm dx\,\mathrm dy\implies E[\mathrm{Var}[X\mid Y]]=\frac12

\mathrm{Var}[E[X\mid Y]]=\mathrm{Var}\left[\dfrac Y2\right]=\dfrac{\mathrm{Var}[Y]}4\implies\mathrm{Var}[E[X\mid Y]]=\dfrac12

\implies\mathrm{Var}[X]=\dfrac12+\dfrac12\implies\boxed{\mathrm{Var}[X]=1}

5 0
3 years ago
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