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svetlana [45]
3 years ago
14

Solve for y x=y^2+4y

Mathematics
1 answer:
Lorico [155]3 years ago
4 0
x=y^2+4y\\ y^2+4y-x=0\\\Delta=4^2-4\cdot1\cdot(-x)=16+4x\\\\
1.\ \Delta0\\
\sqrt{\Delta}=\sqrt{16+4x}=\sqrt{4(4+x)}=2\sqrt{x+4}\\
y_1=\frac{-4-2\sqrt{x+4}}{2\cdot1}=-2-\sqrt{x+4}\\
y_2=\frac{-4+2\sqrt{x+4}}{2\cdot1}=-2+\sqrt{x+4}\\

-----------------------------------------------------

x=y^2+4y\\ y^2+4y-x=0\\
y^2+4y+4-4-x=0\\
(y+2)^2=x+4\\
y+2=\sqrt{x+4} \vee y+2=-\sqrt{x+4}\\
y=-2+\sqrt{x+4} \vee y=-2-\sqrt{x+4}\\

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In which

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e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

The problem states that:

The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and mean 0.2 each day during the weekend.

To find the mean during the time interval, we have to find the weighed mean of calls he receives per day.

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The weekend is 2 days long, with a mean of 0.2 calls per day.

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P(X = 2) = \frac{e^{-0.1286}*0.1286^{2}}{(2)!} = 0.0073

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.

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