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3 years ago

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100 POINTS!!!! Nicholas has some cans at home to donate to the soup kitchen, but decides to start a can drive at his school to s

ee if others will help. The equation representing this is y = 235x + 15, where x is the number of days, and y is the number of cans collected. Which statements are true based on the equation given. Select all that apply. Nicholas had 235 cans at home. The school collected 235 cans a day. Nicholas had 15 cans at home. The school collected 15 cans daily.

2 answers:

gladu [14]3 years ago

8 0

Answer:

he school collected 235 cans a day

Nicholas had 15 cans at home

Step-by-step explanation:

The school collected 235 cans a day

Nicholas had 15 cans at home

Step-by-step explanation:

Looking at the equation:

15 is a constant, meaning that even at x=0 days, nicholas would have had 15 cans, meaning he already had them at home

235 is the number that the variable(the number of days) is multiplied with, this means that 235 is the number of cans that can be collected daily.

Read more on Brainly.com - brainly.com/question/9966828#readmore

docker41 [41]3 years ago

5 0

Answer:

The school collected 235 cans a day

Nicholas had 15 cans at home

Step-by-step explanation:

15 is a constant, meaning that even at x=0 days, nicholas would have had 15 cans, meaning he already had them at home

235 is the number that the variable(the number of days) is multiplied with, this means that 235 is the number of cans that can be collected daily.

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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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4 years ago