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3 years ago

14

one serving of oatmeal provides 16% of fiber you need daily. you must get the remaining 21 grams of fiber from other resources.

how many grams of fiber should you consume daily?

1 answer:

stira [4]3 years ago

3 0

1 - 0.16 = 0.84
21/0.84 = 25
25 grams of fiber should be consumed daily.

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Let X and Y be independent continuous random variables that are uniformly distributed on (0,1). Let H=(X+2)Y. Find the probabili

belka [17]

Looks like there are two distinct problems here.

"Let X and Y be independent ..."

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Conditioning on $X$ involves interpreting this probability for all possible fixed values of $X=x$ :

$P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_xP\left(Y\ge\dfrac{e^z}{X+2}\text{ and }X=x\right)$

$P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_xP\left(Y\ge\dfrac{e^z}{X+2}\mid X=x\right)P(X=x)$

We have $x\in(0,1)$ , and over this domain $P(X=x)=1$ . Since $X=x$ is fixed, we can omit the conditioning notation to leave us with

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$Y$ has CDF

$P(Y\le y)=F_Y(y)=\begin{cases}0&\text{for }y1\end{cases}$

Then we can write our integral as

$P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_0^1\left(1-F_Y\left(\frac{e^z}{x+2}\right)\right)\,\mathrm dx$

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$P\left(Y\ge\dfrac{e^z}{X+2}\right)=x-e^z\ln(x+2)\bigg|_{x=0}^{x=1}$

$P\left(Y\ge\dfrac{e^z}{X+2}\right)=\boxed{1-e^z(\ln2-\ln3)}$

"Let X be a standard normal ..."

Using the method of distribution functions, we have

$F_Z(z)=P(Z\le z)=P(F_X(X)\le z)=P(X\le {F_X}^{-1}(z))=F_X({F_X}^{-1}(z))=z$

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2 years ago

An engineer is designing a play set for an elementary school. The drawing above shows the detailed view for the pre-drilled hole

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3 years ago

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Every 30 minutes. (this was using the LCM method)

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Answer:

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2 years ago

Please help I need help with my hw

olga2289 [7]

Answer:

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Step-by-step explanation:

1 = 5

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$\frac{0.5}{1} :\frac{y}{5}$

y × 1 = 0.5 × 5

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3 years ago