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nydimaria [60]
3 years ago
8

There are 7 red 3 green 2 blue and 8 purple marbles in a bag if a marble is randomly chosen 250 times predict how many times it

should be red or green
Mathematics
1 answer:
kati45 [8]3 years ago
3 0

Answer:

It should be red or green 125 times

Step-by-step explanation:

The first thing to do here is to calculate the probability of selecting a red or a green marble

Total number of marbles = 7 + 3 + 2 + 8 = 20

Probability of selecting a red marble is 7/20

Probability of selecting a green marble is 3/20

The probability of selecting a red or a green marble = Probability of selecting a red marble + Probability of selecting a green marble = 7/20 + 3/20 = 10/20 = 1/2

Now our selection spans 250 times, the number of times it should have been a green or a red marble = The probability of selecting a green or a red marble * number of selection times = 1/2 * 250 = 125 times

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Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{
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y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

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

  • Note 1: A more general solution is y = Ce^x for some constant C.
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Hope it’s right best luck with your studying
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