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bonufazy [111]
3 years ago
13

Factor 81a^2b^4-c^6 completely

Mathematics
1 answer:
olga nikolaevna [1]3 years ago
3 0

Answer: (9ab^2+c^3)(9ab^2-c^3)

Step-by-step explanation:

81a^2b^4-c^6

(9ab^2)^2-(c^3)^2

(9ab^2+c^3)(9ab^2-c^3)

Hope this helps, HAVE A BLESSED AND WONDERFUL DAY! As well as a great Superbowl Weekend! :-)

- Cutiepatutie ☺❀❤

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worty [1.4K]

Answer:  The probability of pulling a red marble is 48%, probability of pulling a blue marble is 32%, probability of pulling a green marble is 20%

and

the probability of pulling three green marbles out with replacement is 0.8%.

Step-by-step explanation:  Given that a bag of marbles contains 12 red marbles 8 blue marbles and 5 green marbles.

Let S be the sample space for the experiment of pulling a marble.

Then, n(S) = 12 + 8 + 5 = 25.

Let, E, F and G represents the events of pulling a red marble, a blue marble and a green marble respectively.

The, n(E) = 12, n(F) = 8  and  n(G) = 5.

Therefore, the probabilities of each of these three events E, F and G will be

P(E)=\dfrac{n(E)}{n(S)}=\dfrac{12}{25}=\dfrac{12}{25}\times100\%=48\%,\\\\\\P(F)=\dfrac{n(F)}{n(S)}=\dfrac{8}{25}=\dfrac{8}{25}\times100\%=32\%,\\\\\\P(G)=\dfrac{n(G)}{n(S)}=\dfrac{5}{25}=\dfrac{5}{25}\times100\%=20\%.

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P(G)\times P(G)\times P(G)=\dfrac{5}{25}\times\dfrac{5}{25}\times \dfrac{5}{25}=\dfrac{1}{125}=\dfrac{1}{125}\times100\%=0.8\%.

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and

the probability of pulling three green marbles out with replacement is 0.8%.

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