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vodomira [7]
3 years ago
12

A culture started with 4,000 bacteria. After 2 hours it grew to 4400 bacteria. Predict how many bacteria will be present after 1

3 hours round your answer to the nearest whole number.
Mathematics
2 answers:
Art [367]3 years ago
5 0
 <span>idk why they want the dratted "k" . it is so simple w/o it ! 
in the form y = ab^x, a = 4000, b = 4400/4000 = 1.1, x = t/2 
N = 4000*1.1^(13/2) = 7432 <------ 
----------------- 
now you won't be satisfied with simplicity, so 
4400 = 4000e^2k 
e^2k = 1.1 
k = ln 1.1/2 = .0477 to 4 dp 
N(13) = 4000e^(13*.0477) = 7436 <------ 

note that apart from a roundabout way, you get a less accurate ans !</span>
Over [174]3 years ago
3 0

Answer:

7436

Step-by-step explanation:

y = ab^x

a = 4000

b = 4400/4000 = 1.1

x = t/2 

4400 = 4000e^2k 

e^2k = 1.1 

k = ln 1.1/2 = .0477 to 4 dp 

N(13) = 4000e^(13*.0477) = 7436


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(a) Find <em>c</em> such that <em>f</em> is a proper density function. This would require

\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}

(b) Get the marginal density of <em>X</em> by integrating the joint density with respect to <em>y</em> :

f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}

(c) Get the marginal density of <em>Y</em> by integrating with respect to <em>x</em> instead:

f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}

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f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}

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E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}

E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}

E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}

(f) Note that the density of <em>X</em> | <em>Y</em> in part (d) identical to the marginal density of <em>X</em> found in (b), so yes, <em>X</em> and <em>Y</em> are indeed independent.

The result in (e) agrees with this conclusion, since E[<em>XY</em>] = E[<em>X</em>] E[<em>Y</em>] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)

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