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LuckyWell [14K]
3 years ago
5

PLEASE HELP MEEEE ILL MARK BRAINLIEST

Mathematics
1 answer:
Gnesinka [82]3 years ago
3 0

DON'T WORRY!

(3x + 2)(3x + 2)(3x - 2)

Notice that (3x + 2) and (3x + 2) are the same, so combining both together becomes this form: (3x + 2)^{2}

And looking at the choices now, notice that the first option is correct if I turn/replace the (3x + 2)(3x + 2)(3x - 2) into (3x + 2)^{2} (3x - 2)

ANSWER: Choose the first option (3x + 2)^{2} (3x - 2)

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3 years ago
Can you please help me out
Taya2010 [7]

The bag contains,

Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,

Total marbles (possible outcome) is,

\text{Total marbles = (R) + (G) +(B) = 9 + 7 + 4 = 20 marbles}

Let P(R) represent the probablity of picking a red marble,

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P(B) represent the probability of picking a blue marble.

Probability , P, is,

\text{Prob, P =}\frac{required\text{ outcome}}{possible\text{ outcome}}\begin{gathered} P(R)=\frac{9}{20} \\ P(G)=\frac{7}{20} \\ P(B)=\frac{4}{20} \end{gathered}

Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,

That means once a marble is drawn, the total marbles (possible outcome) reduces as well,

\begin{gathered} \text{Prob of a red marble P(R) =}\frac{9}{20} \\ \text{Prob of }a\text{ blue marble =}\frac{4}{19} \\ \text{After a marble is selected without replacement, marbles left is 19} \\ \text{Prob of red marble + prob of blue marble = P(R) + P(B) = }\frac{9}{20}+\frac{4}{19}=\frac{251}{380} \\ \text{Hence, the probability is }\frac{251}{380} \end{gathered}

Hence, the best option is G.

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1 year ago
What is the value of 3'5/3'3
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Step-by-step explanation:

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#carryonlearning

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May you Compare v5 and 1.3
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