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agasfer [191]
3 years ago
5

A polar bear rest by a stack of 3000 pounds of fish he has caught. He plans to travel 1000 miles across the arctic to bring as m

any fish as possible home to his family, he can pull a sled that holds up 1000 pound of fish but he must eat 1 pound fish at every mile to keep his energy up. what is the maximum amount of fish in pounds the polar bear can transport across the arctic? how does he do it?
Mathematics
1 answer:
Sloan [31]3 years ago
3 0
I think but im not sure, he would have no fish left because he would eat all of it by the time he gets home

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Calculate how much each size per gram, and circle which gives the best value for money. 250g for $2.30, 400g for &3.40, 1Kg
Bezzdna [24]
To do this, we must set up ratios:
Option One:
\frac{2.30}{250g}  \\  \frac{0.0092}{1g}  \\ 0.0092/g
The first options costs only $0.0092/gram (a fraction of a penny)!
Option Two:
\frac{3.40}{400g}  \\  \frac{0.0085}{1g} \\ 0.0085/g
The second option costs only $0.0085/gram (cheaper than option one)!
Option Three: 
This requires a little more work. First, we have to convert the grams into kilograms. For every 1 kg, there is 1,000 g. Therefore, 1,000g costs $5.65. Next, we set up the ratio as usual:
\frac{5.65}{1000g} \\  \frac{0.00565}{1g}  \\ 0.00565/g
The third option costs $0.00565/gram.

Therefore, option three is the cheapest! 

Hope this helps!
4 0
4 years ago
Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.
Cloud [144]

If F_Y(y) is the cumulative distribution function for Y, then

F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)

Then the probability density function for Y is f_Y(y)={F_Y}'(y):

f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}

The nth moment of Y is

E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy

Let u=\ln y, so that \mathrm du=\frac{\mathrm dy}y and y^n=e^{nu}:

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du

Complete the square in the exponent:

nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du

But \frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2} is exactly the PDF of a normal distribution with mean n and variance 1; in other words, the 0th moment of a random variable U\sim N(n,1):

E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1

so we end up with

E[Y^n]=e^{\frac12n^2}

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ArbitrLikvidat [17]

Answer:

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m_a_m_a [10]
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Find the value of x<br> in the figure below
AlexFokin [52]

Answer:

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Step-by-step explanation:

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Hope it helps and is correct!

6 0
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