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drek231 [11]
3 years ago
5

Jake has a bag of 50 beads, some of which are blue and the remaining are green. Jake randomly pulls out a bead from the bag, rec

ords the color, and replaces it in the bag. Jake has already recorded 14 blue and 6 green beads. Based on these results, what is most likely the number of blue beads in the bag?
{A} 15
{B} 20
{C} 30
{D} 35
Mathematics
2 answers:
miv72 [106K]3 years ago
5 0

Answer:

15

Step-by-step explanation:

14 and 6 added together are 20, there are two sets of 20 and one 10 left over.  that means 6 + 6 + 3 = 15

Your answer is A, 15

lana [24]3 years ago
3 0
It should be D: 35. 14 + 6 = 20. I just multiplied that by two, which equals 40. That means there are 2 sets of 14 and 2 sets of 6. Since there is 10 leftover, I divided both by 2. 7 + 3 = 10. 14 x 2 = 28 + 7 = 35
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Suppose a certain population satisfies the logistic equation given by dP
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Using variable separable method we get

\frac{dP}{10P-P^2}=dt

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\int \frac{dP}{10P-P^2}=\int dt             .... (1)

Using partial fraction

\frac{1}{P(10-P)}=\frac{A}{P}+\frac{B}{(10-P)}

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Using these values the equation (1) can be written as

\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt

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On simplification we get

\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C

\frac{1}{10}(\ln \frac{P}{10-P})=t+C

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Substitute t=0 and P=1 in above equation.

\frac{1}{10}(\ln \frac{1}{10-1})=0+C

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\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}

e^{\ln \frac{P}{10-P}}=e^{10t+\ln \frac{1}{9}}

\frac{P}{10-P}=\frac{1}{9}e^{10t}

Reciprocal it

\dfrac{10-P}{P}=9e^{-10t}

P(t)=\dfrac{10}{1+9e^{-10t}}

The population when t = 3 is

P(3)=\dfrac{10}{1+9e^{-10\cdot 3}}

Using calculator,

P=9.999\approx 10

Therefore, the population when t = 3 is 10.

8 0
2 years ago
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